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ELEMENTARY MATHEMATICAL ANALYSIS 



A SERIES OF MATHEMATICAL TEXTS 

EDITED BY 

EARLE RAYMOND HEDRICK 



THE CALCULUS 

By Ellery Williams Davis and William Charles 
Brenke. 

ANALYTIC GEOMETRY AND ALGEBRA 

By Alexander Ziwet and Louis Allen Hopkins. 

ELEMENTS OF ANALYTIC GEOMETRY 

By Alexander Ziwet and Louis Allen Hopkins. 

PLANE AND SPHERICAL TRIGONOMETRY WITH 
COMPLETE TABLES 
By Alfred Monroe Kenyon and Louis Ingold. 

PLANE AND SPHERICAL TRIGONOMETRY WITH 
BRIEF TABLES 
By Alfred Monroe Kenyon and Louis Ingold. 

ELEMENTARY MATHEMATICAL ANALYSIS 

By John Wesley Young and Frank Millett Morgan. 

COLLEGE ALGEBRA 

By Ernest Brown Skinner. 

PLANE TRIGONOMETRY FOR SCHOOLS AND COL- 
LEGES 
By Alfred Monroe Kenyon and Louis Ingold. 

THE MACMILLAN TABLES 

Prepared under the direction of Earle Katmond Hedrick. 
PLANE GEOMETRY 

By Walter Burton Ford and Charles Ammerman. 
PLANE AND SOLID GEOMETRY 

By Walter Burton Ford and Charles Ammerman. 
SOLID GEOMETRY 

By Walter Burton Ford and Charles Ammerman. 

CONSTRUCTIVE GEOMETRY 

Prepared under the direction of Earle Raymond Hedrick. 
JUNIOR HIGH SCHOOL MATHEMATICS 

By W. L. VoSBURGH and F. W. Gentleman. 



ELEMENTARY MATHEMATICAL 
ANALYSIS 



BY 
JOHN WESLEY YOUNG 

PROFESSOR OP MATHEMATICS, DARTMOUTH COLLEGE 
AND 

FRANK MILLETT MORGAN 

ASSISTANT PROFESSOR OF MATHEMATICS, DARTMOUTH COLLEGE 



THE MACMILLAN COMPANY 
1917 

All rights reserved 






Copyright, 1917, 
By the MACMILLAN COMPANY. 



Set up and electrotyped. Published July, 1917. 
Reprinted September, October, 191 7. 



NoriDoat) }fixtt» 

J. 8. Gushing Co. — Berwick & Smith Co. 

Norwood, Mass., U.S.A. 



PREFACE 

This book aims to present a course suitable for students in 
the first year of our colleges, universities, and technical schools. 
It presupposes on the part of the student only the usual mini- 
mum entrance requirements in elementary algebra and plane 
geometry^ 

The .book has been written with the hope of contributing 
something toward the solution of the problem of increasing 
the value and significance of our freshman courses. The 
recent widespread discussion of this problem has led to the 
general acceptance on the part of many teachers of certain 
principles governing the selection and arrangement of mate- 
rial and the point of view from which it is to be presented. 
Among such principles, which have guided us in the prepara- 
tion of this text, are the following. 

1. More emphasis should be placed on insight and under- 
standing of fundamental conceptions and modes of thought, 
less emphasis on algebraic technique and facility of manipula- 
tion. The development of proficiency in algebraic manipulation 
as such we believe has little general educational value. It is 
valuable only as a means to an end, not as an end in itself. A 
certain amount of skill in algebraic reduction is, of course, 
essential to any effective understanding of mathematical pro- 
cesses, and this minimum of skill the student must secure. 
But it seems undesirable in the first year to emphasize the 
formal aspects of mathematics beyond what is necessary for 
the understanding of mathematical thought. This is espe- 
cially true for that great majority of students who do not 
continue their study of mathematics beyond their freshman 



:i79f40ft 



vi PREFACE 

year and who study mathematics for general cultural and dis- 
ciplinary purposes. It seems to us altogether probable, how- 
ever, that even in the case of students who expect to use 
mathematics in their later life work (as scientists, engineers, 
etc.) greater power will be gained in the same length of time, 
if their first year in college is devoted primarily to the gain- 
ing of insight and appreciation, rather than technical facility. 
Experience has shown only too conclusively that in many cases 
the emphasis usually placed on formal manipulation effectually 
prevents the development of any adequate sort of independent 
power. 

2. The reference above to the general cultural and disciplin- 
ary aims of mathematical study at once raises the question as 
to the selection of the material that is to form the content of 
the course. The cultural motive for the study of mathematics 
is found in the fact that mathematics has played and contin- 
ues to play in increasing measure an important role in human 
progress. An educated man or woman should have some con- 
ception of what mathematics has done and is doing for man- 
kind and some appreciation of its power and beauty. To this 
end our introductory courses should cover as broad a range of 
mathematical concepts and processes as possible. In particu- 
lar, they must not confine themselves to ancient and medieval 
mathematics, but must give due consideration to more modern 
mathematical disciplines. The fundamental conceptions of 
the calculus must be introduced as early as is feasible in view 
of the essential role they have played in the progress of 
civilization. 

If this broad cultural aim is accepted as one of the funda- 
mental principles in the selection of material, we shall readily 
agree that much that is now generally considered necessary 
can and should be eliminated from our general courses in 



PREFACE vii 

mathematics. Almost all of the conventional course in solid 
geometry would fall in this category, as well as much of what 
is now taught as college algebra, all of the more specialized 
portions of analytic geometry, etc. The time thus gained could 
then be used for topics that are culturally more significant. 

3. The disciplinary motive for the study of mathematics 
is the one most often emphasized and ,need not be elaborated 
here. In spite of much recent criticism of the doctrine of 
formal discipline in education and in spite of the fact that 
some of this criticism as applied to mathematics seems to us 
justified, we firmly believe that faith in the disciplinary value 
of mathematics is fundamentally sound. Teachers of mathe- 
matics need, however, to formulate with precision their aims 
and purposes in this direction and make their practice conform 
to this formulation. The disciplinary value of mathematics 
is to be sought primarily in the domain of thinking, reasoning, 
reflection, analysis ; not in the field of memory, nor of skill 
in a highly specialized form of activity. We come back here 
to the conflict between insight and technique discussed earlier 
in this preface. Suflice it to remark here that the purpose 
of technical facility is to economize thought, rather than to 
stimulate it. If our primary purpose is to stimulate thought, 
we must place the major emphasis on the mathematical formu- 
lation of a problem and on the interpretation of the final re- 
sult, rather than on the formal manipulation which forms the 
necessary intermediate step ; on the derivation of a formula 
rather than merely on its formal application ; on the general 
significance of a concept rather than on its specialized function 
in a purely mathematical relation. 

If we desire to enhance the general disciplinary value of 
mathematics, we will seek out and emphasize especially those 
conceptions and those modes of thought of our subject which 



viii PREFACE 

are most general in their application to the problems of our 
everyday life. It is fortunate for our purpose — and it is 
probably more than a mere coincidence — that the conceptions 
and processes of mathematics which most readily suggest 
themselves in this connection are the same that are suggested 
by the cultural motive discussed earlier. The concept of funo- 
tionality and the mathematical processes developed for the 
study of functions are precisely the things in mathematics 
that have most effectively contributed to human progress in 
more modern times ; and the thinking stimulated by this 
concept and these processes is fundamentally similar to the 
thought which we are continually applying to our daily prob- 
lems. "Functional thinking," to use Klein's famous phrase, 
is universal. It comes into play when we make the simplest 
purchase, as well as when we attempt to analyze the most 
complicated interplay of causes and effects. 

In the preparation of this text, we have sought to give an 
introduction to the elementary mathematical functions, the 
concepts connected therewith, the processes necessary to their 
study, and their applications. By making the concept of a 
function fundamental throughout we believe we have gained 
a measure of unity impossible when the year is split up among 
several different subjects. The arrangement of this material 
is exhibited in the table of contents and the text proper, and 
need not be discussed here. We would merely call attention 
briefly to some features which seem to require emphasis or 
explanation. 

The change in the value of a function due to a change in 
the value of a variable is emphasized from the very beginning. 
The change ratio A y/A x is introduced in Chapter III for the 
linear function, and the derivative is introduced as the slope 
of the graph of a quadratic function in Chapter IV, although 



PREFACE ix 

the word "derivative" is not introduced until Chapter XIX. 
Derivatives are used in Chapters IV, V, X, XII, XIII, and XIX. 

We have discussed rather more fully than is customary 
those topics which involve new and important concepts, and 
have been correspondingly brief where we felt the student 
ought to be able to supply the argument himself. We have 
tried throughout to place the emphasis on an understanding 
of the general bearing of the principles, and have consistently 
tried to minimize difficulties of mere algebraic technique. It 
seems quite likely that customary classroom procedure will, 
therefore, need to be modified in the direction of lessening the 
time given to formal recitations and increasing the opportuni- 
ties for informal discussion. A number of questions have 
been inserted among the exercises which it is hoped will 
stimulate such discussion ; this is the purpose also of a num- 
ber of the " Why's " scattered throughout the text. 

The lists of "Miscellaneous Exercises" found at the end 
of chapters beginning with Chapter XI contain some exercises 
too difficult for assignment in an average class. These may 
be used to advantage, we hope, in so-called " honor sections " 
consisting of men who have shown exceptional ability in 
mathematics. 

A word regarding our conception as to how the text may 
be applied to meet the varying mathematical preparation of 
students will not be out of place. At Dartmouth College we 
propose to distinguish in this connection only two kinds of 
freshmen : those who enter without trigonometry, and those 
who have passed a course in trigonometry in their secondary 
school. The former will cover the first fifteen chapters of 
this text in a course meeting three hours per week throughout 
the year (about ninety assignments). These men will have 
all the necessary preliminarj^ training for the usual courses 



X PREFACE 

in the calculus. Those students who enter with trigonometry 
will cover the first nineteen chapters in a course meeting three 
hours per week throughout the year, covering the material of 
Chapters VI, VII, VIII, and IX (which for them is largely 
review) in about three weeks. 

In a course meeting five times per week throughout the 
year, there should be ample time also for a thorough study 
of the important topics of Chapter XX (Determinants) and 
Chapters XXI-XXII (Functions of two independent variables ; 
analytic geometry of space). 

So much has been said in recent years in favor of a unified 
course in mathematics for freshmen that it seems desirable 
actually to try it out in practice. For this purpose a text- 
book is necessary. We do not believe that this text will 
solve the problem ; the most we can hope for is that we have 
secured a first approximation. It is for this reason that we 
urgently request users of this text to communicate to us any 
criticisms or suggestions that occur to them looking to the 
improvement of possible later editions. In particular, we 
should like advice and counsel as to the possible desirability 
of increasing the amount of calculus included in the first year. 
This could be done by devoting less space to the purely geo- 
metric aspects of analytic geometry. On theoretical grounds 
we believe this to be desirable. We felt, however, that we 
ought to be conservative in case of an innovation of this sort, 
with a view of seeing how the introduction to this limited 
extent of the notion of the derivative in the first year fares. 
If the results are satisfactory, we could then take the next step 

with confidence. 

J. W. YOUNG. 
F. M. MORGAN. 
Hanover, N. H., 
AprU, 1917. 



CONTENTS 

PART I. INTRODUCTORY CONCEPTIONS 

Chapter I. Functions and Their Representation 
Chapter II. Algebraic Principles and Their Connection 
with Geometry 



1-32 



Chapter 

Chapter 

I. 

IT. 

III. 

Chapter 

Chapter 

Chapter 

Chapter 



PART II. ELEMENTARY FUNCTIONS 

III. The Linear Function. The Straight Line 

IV. The Quadratic Function . 
Graphs of Quadratic Functions . 
Applications of Quadratic Functions '. 
Quadratic Equations .... 

V. The Cubic Function. The Function jr" 

VI. The Trigonometric Functions 

VII. Trigonometric Relations 

VIII. The Logarithmic and Exponential 

tions 



Chapter IX. Numerical Computation . 

I. Errors in Computation .... 
II. Logarithmic Solution of Triangles 

III. The Logarithmic Scale. The Slide Rule 

IV. Logarithmic Paper .... 
Chapter X. The Implicit Quadratic Functions . 



Func 



64-97 
98-128 

98-114 
115-119 
120-128 
129-142 
143-187 
188-211 

212-235 
236-264 

236-241 
242-251 
252-259 
260-264 



PART III. APPLICATIONS TO GEOMETRY 

Chapter XL The Straight Line 293-319 

Chapter XII. The Circle 

xi 



xii CONTENTS 

Chapter XIII. The Conic Sections . 887-876 

Chapter XIV. Polar Coordinates . . 877-891 

Chapter XV. Parametric Equations 892-401 



PART IV. GENERAL ALGEBRAIC METHODS — 
THE GENERAL POLYNOMIAL FUNCTION 

Chapter XVI. Miscellaneous Algebraic Methods . . 402-419 
Chapter XVII. Permutations, Combinations, and Proba- 
bility. The Binomial Theorem 420-431 

Chapter XVIII. Complex Numbers 482-448 

Chapter XIX. The General Polynomial Function. The 

Theory of Equations .... 449-474 

Chapter XX. Determinants 476-498 

PART V. FUNCTIONS OF TWO VARIABLES 
— SOLID ANALYTIC GEOMETRY 

Chapter XXI. Linear Functions. The Plane and Straight 

Line 494-618 

Chapter XXII. Quadratic Functions. Quadric Surfaces . 614-681 

Tables . . . • 684-642 

Powers and Roots 534 

Important Constants 535 

Four-Place Logarithms 536-537 

Four-Place Trigonometric Functions .... 538-542 

Index 648-648 



ELEMENTARY MATHEMATICAL ANALYSIS 



ELEMENTARY 
MATHEMATICAL ANALYSIS 

PART I. INTRODUCTORY CONCEPTIONS 

CHAPTER I 
FUNCTIONS AND THEIR REPRESENTATION 

1. The General Idea of a Function. Our daily activities 
continually furnish us with examples of things that are related 
to one anotherj of quantities which depend on certain other 
quantities, which change when certain other quantities change. 
Thus, a man's health is related to the food he eats, the exer- 
cise he takes, and to many other things. The price of any 
manufactured article depends on the cost of production, while 
the latter cost in turn depends on the cost of the raw ma- 
terial, the cost of labor, etc. The weather depends on a 
variety of conditions. These are complicated examples of 
dependence. There are very simple examples. Thus the 
price paid for a certain quantity of sugar depends on the num- 
ber of pounds bought and the price per pound ; the area of a 
square depends on the length of one of its sides ; and so forth. 

In all such cases, where some quantity depends on some 
other quantity or quantities, we say that the former is a func- 
tion of the latter. Thus the price of an article is a function of 
the cost of production, the area of a square is a function of the 
length of one of its sides, etc. 

B 1 



2 MATHKMATICAL ANALYSIS [I, § 2 

2. General Laws. Many problems of science consist in 
expressing as accurately as possible one quantity in terms of 
another quantity on which the first depends. The statements, 
" The area of a square is equal to the square of the length of 
one side," and " The speed of a body falling from rest is pro- 
portional to the time it has fallen " are simple examples. 

At the basis of this idea of dependence or functionality is 
the notion of a general law which the quantities in question 
obey. Most of the problems of civilized life are concerned, 
directly or indirectly, with the investigation of such laws. 
Thus medical science seeks to discover the laws governing 
health, economies' investigates the laws governing the produc- 
tion and distribution of wealth, the business man studies the 
conditions which influence his business and his profits. In 
every case the investigation of the law in question involves 
finding out how something is related to, depends on, changes 
with, something else ; i.e. the study of a function of some kind. 

The ability to think clearly about such relationships is of 
the highest importance to every one. This course in mathe- 
matics is primarily concerned with the study of certain of the 
simpler kinds of functions and their applications. 

3. Numbers and Quantities. We shall confine ourselves 
in general to the study of relations between things which can 
be ^measured. We can then always speak of the amount of one 
of them. Such an amount is expressed, in terms of a suitable 
unit of measure^ by means of a number. Anything that can be 
represented by means of a number we shall call a quantity. 

A function expressing the relation of one such quantity to 
another gives rise to a relation between numbers. A very power- 
ful aid in studying functions is their geometric representation, 
which we shall discuss presently. We must consider first, 
however, the geometric representation of a single quantity. 



I, § 5] REPRESENTATION OF FUNCTIONS 3 

4. The Arithmetic Scale. The distinction between two of 
the simplest kinds of quantities can be illustrated by reference 
to their geometric or graphic representation. Every one is 
familiar with the so-called arithmetic scale (Fig. 1), of which the 

I i j I \ 1 1 ^ 1 . , . ^ r- 

J 3 3 4 5 6 

Fig. 1 

yard stick and tape measure are examples. The divisions of 
the scale in these cases represent lengths. Another example 
is the beam on a certain kind of balance ; here the divisions of 
the scale represent weights. 

A characteristic feature of an arithmetic scale is that it 
begins at some point and extends from in one direction. 
The quantities represented by such a scale are expressed by 
means of the numbers of arithmetic. These in turn represent 
simply the magnitude, or the size, or the amount, of something 
(as 12 yd. of cloth, 96 lb. of sugar, etc.). 

5. The Algebraic Scale. Hardly less familiar nowadays is 
the so-called algebraic scale (Fig. 2). The most familiar ex- 

n — ' — I — I — I — ' — r^ — I J , I — ' — r 

-3 -2 •^l/ *1 +2i +3 

Fia. 2 

ample is probably the scale on an ordinary thermometer. 
Every one knows the meaning of -f- 10° or — 5°. 

Such an algebraic scale extends in two opposite directions 
from some arbitrary point (marked 0) of the scale. The quan- 
tities represented by the points of such a scale are expressed 
by means of the so-called real numbers of algebra, such as : 

..., - 4, - Vi2, - 3, - i 0, + 1, + ii, -. 
Such a number represents not merely a magnitude, but 
rather a magnitude and one of two opposite directions or 



4 MATHEMATICAL ANALYSIS [I, § 5 

senses. These two opposite " senses " are of various kinds 
according to tlie quantities considered. They are often ex- 
pressed by such phrases as : " to the right of " and " to the 
left of," " above " and " below," " greater than " and " less 
than," " before " and " after," etc. Thus + 10° of temperature 
means a temperature 10° greater than the arbitrary temperature 
which we have agreed to indicate by 0° ; whereas — 5° means 
a temperature 5° less than the temperature indicated l)y 0°. 
It should be noted that 0° of temperature does not mean the 
absence of temperature. 

6. Magnitudes and Directed Quantities. We have seen 
in the last two sections that a number may represent simply a 
magnitude; or, that a number may represent a magnitude and 
one of two opposite directions. The numbers of arithmetic serve 
the former purpose, the positive and negative numbers of 
algebra serve the latter. Thus the number 5 represents 
simply a magnitude, such as a distance of five miles between 
two stations or a period of time of five hours. The numbers 
-h 5 and — 5 also represent magnitudes of five units ; but 
they represent more than this. They may tell us, for example, 
that a station is five miles east of a certain place denoted by 
and that another station is five miles west of the place denoted 
by 0, respectively ; or that an event took place five hours after 
or five hours before a certain event. 

We may then distinguish two kinds of quantities : (1) mag- 
nitudes, and (2) so-called directed quaiitities. Examples of the 
former are : the length of a board, the weight of a barrel of 
flour, the duration of a period of time, the price of a loaf 
of bread, etc. Examples of the latter are : the temperature (a 
certain number of degrees above or below zero), the distance 
and direction of some point yl on a line from some other 
point B on the line^ the time at which a certain event 



, , ,5 , 




. - """^ , ■ - 


< . ."^. 





I, § 7] REPRESENTATION OF FUNCTIONS 5 

occurred (a certain number of hours before or after a given 
instant) ; etc.* 

Geometrically, the distinction between directed quantities 
and mere magnitudes corresponds to the fact that, on the one 
hand, we may think of the line segment AB as drawn from A to 
B or from B to A, and, on the other hand, we 
may choose to consider only the length of 
such a segment, irrespective of its direction. 
Figure 3 exhibits the geometric representation 
of 5, 4- 5, and — 5. A segment whose direc- 
tion is definitely taken account of is called a directed segment. 
The magnitude of a directed quantity is called its absolute 
value. Thus the absolute value of — 5 (and also of -h 5) is 5. 

7. Further Remarks concerning Scales. Scales, both arith- 
metic and algebraic, occur in practice in a variety of forms. We have 
hitherto considered only the simplest form, in M^hich the scale is con- 
structed on a straight line and in which the subdivisions corresponding 
to the numbers 1, 2, 3, ... (and in case of the algebraic scale also those 
corresponding to the numbers — 1, —2, —3, •••) are at equal intervals. 
Neither of these two conditions is essential. A scale may be constructed 
on a curved line (a circle, for example, in which case it is sometimes 
called a dial). Scales are also used in which the intervals between the 
points representing the whole numbers are not equal. Such a scale is 
called a non-uniform scale. The scales on some forms of thermometers, 
on a slide rule (see p. 252), on certain types of ammeters and pressure 
gauges, etc., may serve as examples of non-uniform scales. The scales 
discussed in §§ 4, 5 are then to be described more fully as rectilinear and 
uniform. In the future, unless specifically stated otherwise, a scale will 
always mean a uniform scale. 

* We are here considering only magnitudes in one of two opposite directions 
It is also possible to consider as quantities magnitudes taken in any direction 
in a plane or in space. Thus a force has a certain magnitude 
and is exerted in a certain direction ; it could be completely 
represented by a line segment whose length represents the 
magnitude of the force and whose direction (shown by arrow- 
head) represents the direction in which it acts. Such quantities are called 
vectors. We shall have occasion to refer to them again (Chap. XVIII) . 



6 



MATHEMATICAL ANALYSIS 



[I, §8 



8. Use of Line Segments to Represent Quantities. Statis- 
tical Data. A common use of line segments to represent 
quantities is in connection with the graphic representation of 
statistical data. The table below, for example, gives the areas 
of the New England States ; the adjacent figure represents 
these areas by means of line segments. 

Area of New England States 



States 


Maine 


Vermont 


New 
Hampshire 


Massa- 
chusetts 


Connec- 
ticut 


Rhode 
Island 


Square Miles 


33,040 


9,565 


9,305 


8,315 


4,990 


1,250 



Maine 
Vermont 
New Hamp, 
Massachusetts 
Connecticut 
Rhode Island 



JO 



15 

Fig. 4 



?o 



25 



30 Thous.sq.miles 



The method of constructing such a graphic representation 
should be clear without further comment. 

The above areas could also be represented by areas, as in the following 
figure. 




Vt. 



N.H. 



Mass. 



E3 



Fig. 5 
In general, this method of representation is not so serviceable. Why ? 



I, § 8] REPRESENTATION OF FUNCTIONS 7 

EXERCISES 

1. From the following table represent graphically by means of line 
segments the enrolment in Dartmouth College during the years 1901-1916 : 

'01-'02 '02-'03 '03-'04 '04-'05 '05-'06 '06-'07 "07-'08 '08-'09 

686 709 802 857 927 1058 1131 1136 

'09-'10 'lO-'ll '11-'12 '12-'13 13-' 14 '14-' 15 '15-'16 

1197 1165 1242 1246 1284 1336 1422 

Use a convenient unit to represent 100 students (say \ in.). Can you 
then represent the data with complete accuracy ? Why ? 

2. Represent graphically the size of the libraries of the following 

institutions : 

No. of Volumes No. of Volumes 

Harvard 1,180,000 Williams 80,000 

Yale 1,000,000 Amherst 110,000 

Dartmouth 130,000 Wesleyan 100,000 

Brown 115,000 Univ. of Vermont . . 91,000 

3. Take the edge of a sheet of paper and mark on it a point A. Place 
this edge along the segment representing the area of Vt. in Fig. 4, the 
point A coinciding with the left-hand extremity of the segment. Mark 
the right-hand extremity by a point B on the paper. Do the same with 
the segment representing N. H., placing the point B at the left-hand 
extremity, however, and obtaining a new point C, corresponding to the 
right-hand extremity. Continue this process for the states Mass., Conn., 
and R. I. The total segment represents the sum of the areas. Show that 
Me. has an area almost as great as that of the other N. E. states com- 
bined. The process just described in the above exercise is known as 
graphic addition. 

4. Describe a similar process for graphic subtraction. 

5. Show that the distance between two points of an arithmetic scale 
can always be found by subtraction. Is the same true for the points of 
an algebraic scale ? What is the meaning of the sign of the difference ? 

6. Two algebraic scales intersect at right angles, the point of intersec- 
tion being the point of each scale, and the units on both scales being 
the same. Show how to find the distance from any point on one scale to 
any point on the other. Would your method still be applicable, if the 
units on the two scales were different ? Explain your answer. 

7. In constructing Fig. 5 what theorem of plane geometry regarding 
the areas of similar figures is used ? Could the result of Ex. 3 have been 
readily obtained from the representation in Fig. 5 ? 



8 



MATHEMATICAL ANALYSIS 



[I, §9 



9. The Investigation of Functions. We are now ready 
to consider in some detail a few special examples of func- 
tions, in order to familiarize ourselves with certain gen- 
eral characteristics a function may possess, with certain 
methods for the representation and study of functions, and 
with the terminology. This is desirable before taking up the 
more systematic study of general types of functions. 

10. Example i. Tlie temperature as a function of the time. 
The temperature at a given place is a function of the time of 
day. At any given time we can determine the temperature by 

/ 7 'Wednesday I S TJmrsday J 9 Fn 1<j;/ 




Fig. 6 

simply reading an ordinary thermometer. For the meteorolo- 
gist, however, the actual temperature at any instant is of less 
importance than the changes in the temperature that take place 
during a period of time (such as a day, a month, etc.). To 
trace these changes he must know the temperature at every 



I, § 10] REPRESENTATION OF FUNCTIONS 



9 



instant. For this purpose he makes use of a self-recording 

thermometer. A portion of a record of such a thermometer is 

given in Fig. 6. 

The way in which such an instrument works is briefly as follows. 
The pivoted lever shown in the figure (Fig. 7) carries a pencil point. The 
mechanism of the instrument causes the pencil end of the lever to rise or 
fall as the temperature rises or falls, so that if a vertical thermometer 
scale* were adjusted behind the pencil point we could read off the 




Fig. 7 



temperature on this scale. The pencil point rests against a strip of paper, 
ruled as in Fig. 6, which is mounted on a drum. Clockwork causes this 
drum to rotate uniformly at the proper speed. The rulings on the strip 
of paper now explain themselves. The distance between two successive 
horizontal lines corresponds to 2° of temperature. The distance between 
two successive vertical arcs corresponds to two hours. The temperature 
at any instant can then be read from the record on the strip of paper. 

The way in which such a record may be used is illustrated 

by the following questions, which refer to the record of Fig. 6. 

* Since the pencil moves on an arc of a circle, this vertical scale is con- 
veniently constructed on such an arc, rather than on a straight line. 



10 MATHEMATICAL ANALYSIS [I, § 10 

1. What was the temperature at noon on each of the three days given ? 

2. What was the temperature at midnight between Wednesday and 
Thursday ? At 6 p.m. on Friday ? 

3. What was the maximum and the minimum temperature on each of 
the three days, and at what times did it occur ? 

4. When was the temperature 50° ? During what periods was it above 
50°? 

5. How would a stationary temperature be recorded ? A rapidly 
rising temperature ? A rapidly falling temperature ? 

6. By how many degrees did the temperature change on Wednesday 
from noon to 2 p.m. ? Was this change a rise or a fall ? 

7. During what two hours on these three days did the greatest rise in 
temperature occur ? 

8. When did the most rapid rise in temperature occur ? When the 
most rapid fall ? 

9. What was the average rate of increase (in degrees per hour) in the 
temperature from the minimum on Thursday to the maximum on Thurs- 
day ? The average rate of decrease from the maximum on Wednesday to 
the following minimum ? 

11. Graphic Representation. In the preceding example we 
exhibited the temperature as a function of the time by means 
of a curve drawn with reference to a time scale and a tempera- 
ture scale. Such a curve is called a graph of the function in 
question. Such a graphic representation gives a vivid picture 
of the function ; but it is limited in accuracy. Why ? Can a 
change in temperature of 0.1° be distinguished on this graph ? 

12. Example 2. Speed in terms of the time. Keadings of 
the speedometer of an automobile taken every five seconds 
from a standing start are given in the following table : 

Number of seconds after start 5 10. 15 20 25 30 35 
Speed in miles per hour 2 6 7 16 21 28 36 

We proceed to construct a graph of the function thus ob- 
tained, as follows. We take a piece of square-ruled paper and 
on one of the horizontal lines (which for convenience we draw 
more heavily) construct a uniform scale to represent the time 



I, § 12] REPRESENTATION OF FUNCTIONS 



11 



(Fig. 8). On the vertical lines through the points representing 
5, 10, 15, 20, . . . seconds we lay off segments to represent the 
speeds at the respective instants. This is most conveniently- 
done by constructing on the vertical line through a scale 
representing speed in miles per hour. Thus, by reference to 
the scale indicated in the figure, the point A represents the 



«y 


■ ~ 






- 




r 




~ 




~ 






■" 




~ 


~ 




~ 


— 


" 


- 












~ 




" 




" 




" 






■ 




























































































































































































































































































































^ 






























































































































































/ 






























































































































































f 














































































^ 






























































































































































1/ 












































































































































































































































/ 














































































/ 












































































































































































































































/ 






























</n 
















































4 
































• W 














































, 


























































































































































































































































































































/ 














































































/- 






























































































































































f 






























































































































































f 














































































J 












































































t 






















































































































































1 




-i 




* 
































































































































































































































. 














































































•X 


' 










































































^ 


' 










































































:: 


















J 




















































_ 







10 no i 

Seconds after start 

Speed of an Automobile 

Fig. 8 



corresponding values : 15 seconds and 7 miles per hour. The 
other points indicated in the figure are now readily located, or 
" plotted," in similar fashion. The final step in constructing the 
figure consists in drawing a " smooth curve " through the points. 
The curve thus obtained may be used as was the tempera- 
ture curve discussed in the previous example. We might, for 
example, conclude from this figure that the speed of the car at 
the end of 23 seconds was probably about 18 ^ miles per hour. 



12 MATHEMATICAL ANALYSIS [I, § 12 

The necessity of saying " probably ", however, exhibits an 
essential difference between this example and the former one. 
In case of the temperature record the temperature at every 
instant was automatically recorded ; any point of the curve in 
that example was as significant as any other point. In the 
present example the only speeds actually measured are those 
specifically listed in the above table. And yet the conclusion 
stated above regarding the speed of the car at the end of 23 
seconds is justified. Why ? 

1. What was the probable speed of the car at the end of 27 seconds? 

2. How long did it take the car to pick up from to 30 miles per hour ? 

3. The driver probably shifted gears between the 10th and 15th seconds. 
What can be said of the reliability of the curve during this interval ? 

4. How is the steepness of the curve related to the rate at which the 
speed is increasing ? 

6. Is it possible to calculate, by the use of this figure, approximately 
how far the car traveled during the first 35 seconds ? 

13. Variables. It is desirable to introduce at this point a 
certain terminology. In the preceding examples we have 
considered temperature and speed as functions of (i.e. de- 
pendent on) the time. We have considered several different 
instants of time and the corresponding values of the tem- 
perature and the speed. Whenever, in a given discussion, 
we consider a number of different values of a quantity, 
such as time, or temperature, or distance, or weight, etc., 
we call such a quantity a variable. In the above examples, 
the time and the temperature and the speed are all varia- 
bles ; and, since in the first example we have thought of 
the temperature as depending on the time, we may speak 
of the temperature as the dependent variable, of the time 
as the independe7it variable. It is often more convenient, 
however, to call the dependent variable simply the function 



I, § 15] REPRESENTATION OF FUNCTIONS 



13 



and the independent variable the variable. Thus, in the 
second example, the speed was the function and the time 
was the variable. 

14. Tabular Representation. Interpolation. In the second 
example we secured data concerning a function by measurement 
and exhibited the corresponding values of variable and function 
by means of a table of values. Such a table is called a tabular 
representation of the function. The accuracy of such a repre- 
sentation is limited only by the precision of measurement. 
Such a table, however, gives an incomplete description of the 
function. AVhy? The process of obtaining values of the 
function for values of the variable that lie between the re- 
corded values stated in the table is called interpolation. 
When the interpolated values are read from a graph of the 
function, the process is known as graphic interpolation. The 
answers to the first two questions at the end of § 12 were 
obtained by graphic interpolation. 

15. Example 3. Volume of ivater as a function of the tem- 
perature. When 1000 cc. of water at 0° centigrade is heated, 
it is found that the volume of the water changes according to 
the following table. 



Degrees Centigrade 
Cubic Centimeters 



1000.00 


2 
999.90 


4 

999.87 


6 
999.90 


8 
999.98 


Degi-ees Centigrade 
Cubic Centimeters 


10 
1000.12 


12 
1000.32 


14 

1000.57 


16 
1000.86 


20 
1001.61 



It requires a rather careful examination of this table to learn 
that as the temperature (the variable) is increased 'from 0° 
the volume of the water (the function) decreases and then in- 
creases. A graphic representation of this function, analogous 
to the examples already considered in §§ 10, 12, would have 
yielded this result at a glance. It is our next concern to 



14 



MATHEMATICAL ANALYSIS 



[I, § 15 



see how such a representation can be constructed, in this 
case. 

To this end we a take a piece of square-ruled paper and on 
one of the horizontal lines construct a uniform scale to repre- 
sent temperatures. At the points representing 0°, 2°, 4°, 6°, 
•••, we would then lay off on the vertical lines distances that are 



















1 1 








-^ 


- 


- 


' 






- 










r- 






r 


r 


■^ 








































































































































































f 




























































y 






























































/ 




























































1 
























JJ 






































/ 




























































/ 


























1001.0 




































f 


























































/ 




























































> 


' 




























































/ 




























































/ 




























































/ 






























































/ 




























































/ 




























































/ 




























































/ 








































1000.0 


r 
















^ 


/' 












































s 


«v, 












^ 


f 
















































9 


^-m 


^ 






s 
























































1 










































































































































































































































































































































































































































999.0 































































^ ""* FlO 9 ^ Degrees Centigrade, 

to represent the volumes in which we are interested. 
However, at this point a difficulty presents itself. The 
numbers representing the volumes in question are so large, and 
the differences between the volumes for the various tempera- 
tures so small, that, if we choose the unit on the vertical 
scale small enough to represent these volumes on a sheet of 
paper of convenient size, it would be a practical impossibility 



I, § 15] REPRESENTATION OF FUNCTIONS 15 

to represent the volumes with sufficient accuracy to make the 
diiferences in the volumes distinguishable. It is precisely 
these variations in volume, however, in which we are primarily 
interested. 

To overcome this difficulty, we adopt the expedient of ex- 
hibiting merely that portion of the graphic representation in 
which we are primarily interested, and are then able to use a 
largely magnified scale. That is, we observe that all the 
volumes in which we are interested lie between 999.00 cc. and 
1001.00 cc. We may then assume that the points on the line 
on which we constructed the temperature scale are at a height 
representing 999.00 cc. (Fig. 9). In other words we suppose 
the zero point' of the vertical volume scale to be a great dis- 
tance below the point at which we are working. We construct 
a portion of the volume scale on the vertical line through 0, 
marking the latter point 999.0 and choosing the unit on this 
scale sufficiently large to meet our requirements. In the 
figure, as drawn, each of the vertical divisions represents 
0.1 cc. The construction of the points P, Q, R, ••• is then 
readily made. A smooth curve drawn through the points thus 
plotted then gives the graph *of the function. 

Here, again, the points in the curve between the points 
given by the table are uncertain ; but the regularity with 
which the given points are arranged together with the nature 
of the phenomenon we are considering leaves little room for 
doubt that, if the volumes for 1°, 3°, 5°, — should be measured 
and the resulting volumes plotted, the resulting points would 
be located upon (or at least very near to) the curve drawn. 

1; What is the volume of water at 1° ? at 19° ? 

2. What is the minimum volume, and at what temperature does it 
occur ? 

3. At what temperature besides 0° is the volume 1000.00 cc? 



16 



MATHEMATICAL ANALYSIS 



[I, § 15 



EXERCISES 

1. The following temperatures were observed at Hanover, N.H., on a 
certain day in February, 1914 : 



Midnight 


-12°F. 










1 A.M. 


-13° 


9 a.m. 


- 12° F. 


6 P.M. 


+ 18° F. 


2 A.M. 


-14° 


10 A.M. 


- 2° 


6 p.m. 


+ 11° 


3 a.m. 


-15° 


11 A.M. 


+ 4° 


7 P.M. 


+ 6° 


4 A.M. 


- 17° 


Noon 


+ 10° 


8 p.m. 


+ 2° 


5 A.M. 


-20° 


1 P.M. 


+ 12° 


9 P.M. 


+ 1° 


6 A.M. 


-21° 


2 P.M. 


+ 14° 


10 p.m. 


0° 


7 A.M. 


-22° 


3 p.m. 


+ 19° 


11 p.m. 


- 2° 


8 a.m. 


-19° 


4 p.m. 


+ 22° 


Midnight 


-' 4° 



Plot the corresponding points on square-ruled paper, and draw an 
approximate graph of the function. Assuming this graph to be correct, 
what was the temperature at 6.30 a.m.? At 6.30 p.m.? What was the 
total range (the difference between the maximum and the minimum) 
of temperature during the day ? How long did it take the temperature 
to rise from its minimum to its maximum ? At what average rate in 
degrees per hour did the temperature rise during this period ? 

2. A stiff wire spring under tension is found experimentally to stretch an 
amount d under a tension T as follows : 



r in lb 


10 


15 


20 


25 


30 


d in thousandths of in. . 


8 . 


12 


16.3 '^ 


20 


23.5 



Plot the above data. What would the stretch be when the tension is 
12 1b.? 271b.? 23 1b.? 

3. The intercollegiate track records are as follows, where d is the dis- 
tance run and t is the time : 



d 


100 yd. 


220 yd. 


440 yd. 


880 yd. 


1 mile 


2 miles 


t' 


9^ sec. 


2H sec. 


48 sec. 


1 m. b^ sec. 


4 m. 14f sec. 


9 m. 23| sec. 



I, § 15] REPRESENTATION OF FUNCTIONS 



17 



Plot these records by points in a plane, and draw a smooth curve through 
them. Are the points of this curve significant ? Why ? What would 
you expect the record for 600 yd. to be ? For 1600 yd.? For 1000 yd.? 
Compare the results of these interpolations with the actual records for 
these distances. 

4. The following table shows the distance at which objects at sea- 
level are visible from certain elevations : 



Elevation 


Distance 


Elevation 


Distance 


Elevation 


Distance 


Feet 


Miles 


Feet 


Miles 


Feet 


Miles 


1 


1.3 


40 


8.4 


200 


18.7 


5 


3.0 


50 


9.3 


300 


22.9 


10 


4.2 


100 


13.2 


500 


29.6 


20 


5.9 


150 


16.2 


1000 


33.4 


30 


7.2 











Plot the graph of this function. Use a different scale for elevation for 
values from 100 to 1000 ft. from that used from 1 to 50 ft. Why ? 

5. The following is an extract of the mortality table prescribed by 
statute in most states as the basis on which the reserves of life insurance 
companies shall be computed : 



Age 


Number 
Living 


Age 


Number 
Living 


Age 


Number 
Living 


10 
15 
20 
26 
* 30 
35 


100,000 
96,285 
92,637 
89,032 
85,441 
81,822 


40 
45 
50 
55 
60 
65 


78,106 
74,173 
69,804 
64,563 
57,917 
49,341 


70 
75 
80 
85 
90 
95 


38,569 

26,237 

14,474 

5,485 

847 

3 



Draw the mortality curve. Of 100,000 living at the age of 10 years 
approximately how many would be alive at 32 years ? At 57 years ? 
How would you represent on the graph the number dying during any 
given period of five years ? 



18 



MATHEMATICAL ANALYSIS 



[I, § 16 



16. Empirical Functions and Arbitrary Functions. The 

examples of functions we have hitherto considered have been 
taken from observed measurements of relations existing in 
nature and life about us. Such functions are called empirical. 
Another type of functions may now engage our attention. 
They may be called arbitrary or artfjicial The following will 
serve as an example. 

17. Example 4. Letter postage. According to the postal 
regulations the postage on letters is fixed at two cents per 



















u 
u 

10 

s 
e 

: 























































































2 3 4 

Letter Postage 
Fig. 10 



G Ounces 



ounce or fraction thereof. The graph showing the relation 
between the amount of postage and the weight of the letter 
is then given by Figure 10. 

18. Constant Functions. Continuous and Discontinuous 
Functions. The graph just referred to exhibits two peculiar- 
ities that we have not yet had occasion to observe in connection 
with a function. 

(1) The value of the function may make a sudden jump as 
the variable passes through certain values (in this case when 
the weight passes through the values 1 oz., 2 oz., etc.) without 
taking on the intermediate values. In the present case, as the 
weight is increased from exactly 1 oz. to the slightest amount 



I, § 19] REPRESENTATION OF FUNCTIONS 19 

above 1 oz. the postage jumps from 2 cents to 4 cents. A 
function with such breaks, or changes of a definite amount 
for no matter how slight a change in the variable, is said to be 
discontinuous for those values of the variable at which the 
break or jump occurs. 

A function, on the other hand, whose graph is a continuous 
line or curve without such sudden breaks or changes is said to 
be a continuous function.* 

(2) Portions of this graph are horizontal straight lines, which 
means that certain changes in the variable produce no corre- 
sponding change in the value of the function. Thus, the 
postage does not change as the weight of the letter is in- 
creased from slightly more than 1 oz. to 2 oz. In such a case 
we say that the function is constant (or stationary) for the 
interval of the variable in question. 

We should observe, further, that the graph of the function 
as drawn does not furnish a unique value for the function at 
the points of discontinuity, i.e. when the weight is 1, 2, 3, ••• 
oz., since there is nothing to indicate whether we should take 
the lower or the higher value. As a matter of fact the arbi- 
trary definition of the function specifies that the lower value is 
to be taken. 

19. More about Arbitrary Functions. We must not assume, 
of course, from the preceding example that every arbitrary 
function is discontinuous. 

In fact, we should note that if we take any square-ruled 
paper, construct on it a horizontal scale, any number of which 
we will designate by x, and a vertical scale, any number of 

* The word continuous is used in mathematics in a highly technical sense, 
the full discussion of which is beyond the scope of an elementary course. 
The definition of the term given above is sufficiently precise for our present 
purposes. Later we shall have more to say of it. 



20 



MATHEMATICAL ANALYSIS 



[I, § 19 



which we will call y, and then draw an arbitrary curve across 
the paper, as in Fig. 11, we thereby define a relation between 
the numbers x of the horizontal scale and the numbers y of 
the vertical scale, such that to every value of x corresponds a 
certain value (or possibly a set of values) of y ; i.e. we define 
y as a function of x.* The .reason for the phrase in paren- 








5eii 




- 












... 




-- ■ w 




- 






" " ' t 


'-'- 


- 




:: 


'^ 


: 


ft 


:- 


±:i 




L 
r 


. ! ■, 






iij 


• 1 


is , . 


" "'4'" 


























. . 


. . . . • 




. 


::: 


i:: 




-^\'.:::i 


: 


5 


T- 


i^^ij 


>4 --■' 


: 


ir 


T^l 


ii 


M 


- 


±i 




; 



Fig. 11 



Fig. 12 



theses in the last sentence is as follows. If the curve we draw 
is such that for any value of x the corresponding vertical 
line cuts the curve in more than one point, there will be 
associated with such a value of x more than one value of y 
(Fig. 12). The variable y is in such a case still a function of 
X, since the values of y are determined by the values of x. 
The distinction between functions of the latter type and those 
previously considered is made by the following definitions : 

If to every value of the variable under consideration there 
corresponds a single value of the function, the function is said 
to be single-valued or one-valued. If to any value of the vari- 
able corresponds more than one value of the function, the 
latter is said to be multiple-valued. 



*The accuracy with which a function is defined by its grapb depends on 
th6 accuracy with which it is possible to read the two scales of reference and 
the ** fineness " of the curve. 



I, § 20] REPRESENTATION OF FUNCTIONS 



21 



We shall for the present be concerned primarily with one- 
valued functions only, although one example of a two-valued 
function will occur soon. Multiple-valued functions will be 
considered later (Chapter X). 



EXERCISES 

1. From the following data construct a graph showing the cost of 
domestic money orders in the United States : 



Amount of Order 


Kate 


Amount of Okder 


Katk, 


• 

Not over $ 2.50 


3 cents 


Over $30.00 to | 40.00 


15 cents 


Over .| 2.50 to |5.00 


5 cents 


Over 40.00 to 50 00 


18 cent^ 


Over 5.00 to 10.00 


8 cents 


Over 50.00 to 60.00 


20 cents 


Over 10.00 to 20.00 


10 cents 


Over 60.00 to 75.00 


25 cents 


Over 20.00 to 30.00 


12 cents 


Over 75.00 to 100.00 


30 cents 



2. Draw a figure showing the rates for parcel-post packages for zone 
1 ; for zone 2 ; for zone 3. Compare these graphs. 

3. Draw a figure to represent the cost of gas in your own city. Is 
there a different rate for large consumers ? If so, will this show clearly 
on the graph ? How ? 

4. On a piece of square-ruled paper draw graphs of continuous func- 
tions which are rapidly increasing ; rapidly decreasing ; slowly increas- 
ing ; slowly decreasing. 

5. Draw the graph of an arbitrary function which is increasing and in 
which the rate at which it increases is increasing. Also that of an in- 
creasing function in which the rate of increase is decreasing. 

20. Analytic Representation of Functions. We have 
hitherto considered two methods of representing a function, the 
graphic and the tabular. There is a third method, called the 
analytic, which in its simplest form consists of the expression 
of the function in terms of the variable by means of d, formula, 
from which the corresponding values of the variable and the 
function can be computed. The following will serve as examples. 



22 



MATHEMATICAL ANALYSIS 



[I, § 21 



21. Example 5. Capital and interest. The amount A in t years 
of $ 1000 drawing simple interest of 5 % is given by the formula 

(1) ^ = 1000 + 50<. 
By substituting for t a suc- 
cession of values and com- 
puting the corresponding 
values of A^ we obtain from 
this formula a tabular rep- 
resentation of the function. 
This in turn can be repre- 
sented graphically. The 



a 

1300 
1250 
1200 
1150 
1100 
1050 
1000 





p 


n 






-r-i 


n 






- 


-1 








- 


- 




- 


- 


■ 




- 


- 






" 




■ 


" 


~ 




- 


' 


" 




' 




- 


- 






- 


■■ 
























































































































































- 






































































































































































































- 








■ 




















































^ 


' 


































































=: 




































































' 


































































^1 




































































» 




























■ 






- 
































, 


•f 








































































































































































































































































































































































































■ 



















































































































































-' 














































































/ 








< 










. 










i 










V 








6 






Y 


e 


n 


ra 





t (years) 



A (dollars) 



1000 



1050 



1100 



1150 



1200 



1250 1300 



table above and Fig. 13 are the 
result.* The points plotted ap- 
pear to be on a straight line. 
Prove that they are. 

22. Example 6. TJie 
area of a square. The area 
(in square inches) of a square 
whose side is x inches long is 
given by the formula 

y = x2. 
From this equation, we readily 
compute the following table. 



^ 



Jl 


m -t -. 


16 /- 


t 


7 


19 ~J- 


12 2 




"F 


5? y _ _ _ 


^ s t - - ~ 


y 


A_ 


, Z 


^ 4- 


7 


.<^ 


0^ 1 2 s 4 '0 



Inches 
Fig. 14 



X (in.) . . 





0.5 


1.0 


1.5 


2.0 


2.5 


3.0 


3.5 


4.0 


y (sq. in.). 





0.25 


1.00 


2.25 


4.00 


6.25 


9.00 


12.25 


10.00 



* In practice bankers do not take account of fractions of a day in comput- 
ing interest. Strictly speaking, therefore, the graph of the function A, as 
used in practice, is discontinuous. This practice of bankers is, however, dic- 
tated by convenience. It does not alter the fact that the function, as such, is 
continuous. 



1, § 24] REPRESENTATION OF FUNCTIONS 



23 



Using these values, it is now easy to draw the graph, which is shown 
in Fig. 14.* 



X 



M. 



*iL 



Fig. 15 



23. Example 7. The function de- 
fined by a circle. It is often desirable 
to obtain an analytic representation of a 
function, originally given graphically or 
by means of a table. Such an analytic 
representation is sometimes easy to obtain. 
Suppose, for example, that on square-ruled 
paper an a:-scale and a ?/-scale have been 
constructed with the units on the two 
scales equals and suppose that with the 
common 0-point of the scales as a center 

a circle is drawn with a radius of 2 units (Fig. 15). The functional rela- 
tion between the variables x and y defined by this curve is to be expressed 
by means of a formula. 

If P is any point on the circle, the absolute values of the x and the y of 
this point form the legs "of a right-angled triangle of which the hypotenuse 
measures 2 units. By a well-known theorem of geometry we have then 

y2 = 4 — a;2 or 

y =± V4 — x^. 

This is the analytic representation sought. It may be noted that we 
have here to do with a two-valued function. 

24. Range of a Variable. We had occasion some time ago 
(§ 13) to introduce the term variable. In the future such a 
quantity will generally be represented by a symbol, such as a;, 
or 2/, or t, etc. Indeed this was done in some of the preceding 
examples. The various values attached to such a symbol 
throughout the discussion are numbers. These numbers con- 
stitute the range of the variable in question. 

The range of a variable is usually determined by the nature 
of the problem under consideration. Often it is very definitely 
restricted. Thus in the case discussed in the last article the 

* When, as here, the only fractional parts of a unit which occur are halves, 
quarters, etc., it is convenient to use a ruled paper on which the larger units 
are subdivided into four or eight parts instead of ten. 



24 MATHEMATICAL ANALYSIS [I, § 24 

range of x (as well as that of y) consists of all (real) numbers 
from — 2 to + 2, and no others. For numbers outside this 
range, the function in question is not defined. Again, in the 
case of the mortality table considered in Ex. 5, p. 17, the range 
of the dependent variable (the number of persons living at a 
given age) is restricted to whole numbers less than 100,000 ; 
fractional values of the variable are here meaningless. 

25. Increasing and Decreasing Functions. A function 
which increases when the variable increases is called an in- 
creasing function ; if, on the other hand, the function decreases 
as the variable increases, the function is called decreasing. 
Thus the amount A of capital and interest recently considered 
is an increasing function of the time t, throughout the range of 
the latter. Also, the area of a square is an increasing function 
of the length of one of its sides. On the other hand, the num- 
ber of people livhig at a given age is a decreasing function of 
the age. A function may be increasing for certain values of 
the variable and decreasing for certain other values. Thus, 
the temperature is during certain portions of the day an 
increasing function, during other portions a decreasing func- 
tion. The volume considered in § 15 is a decreasing function 
of the temperature T, from T = to T = 4, and an increasing 
function for values of T greater than 4.* 

If the two scales with reference to which the grapli of a function is 
constructed are jjlaced in the more usual way, so that the numbers on the 
scales increase to the right and upward, respectively, what distinguishes 
the graph of an increasing function from that of a decreasing one ? 

* In the case of the circle discussed in § 23, the function has two " branches " 
in the interval from x = — 2 tox=4-2, the one consisting of the positive 
values of y, the other of the negative values of y. The function may be con- 
sidered as consisting of two one-valued functions, one of which increases from 
a;= — 2 to a; = and decreases from x = to x=-\-2, while the other de- 
creases from x=— 2 to a; = and increases from a; = to x=-\-2. 



1, § 25] REPRESENTATION OF FUNCTIONS 25 

EXERCISES 

1. If a body falls from rest, its speed v in feet per second at the end 
of t seconds is given by the relation v = 32 t. Construct the graph of v as 
a function of t. 

2. The charge for printing n hundred circulars of a certain kind is 
|) = 2 n + 10 dollars. Represent the function graphically. 

3. The express rate r on a package is computed from the following 

formula : r =-^(p — 30)+ 30, where w is the weight of the package in 
100 

pounds and^ is the charge per hundred pounds. Draw the graph of r as 
a function of w, for each of the values p = 40, 60, 80, 100. What com- 
ment would you make on this rule for p = 30, or for values of p less than 
30 ? This is an example in which the range of the variable is arbitrarily 
limited to be not less than a certain amount. The formula in this exer- 
cise really gives r as a function of the two variables w and p. 

4. When a body is dropped from a height of 200 ft., its distance s 
from the ground at the end of t sec. is given by s = 200 — 16.1 1^. Draw 
the graph of s as a function of t. In how many seconds will the body 
reach the ground ? At what time is the speed of the body greatest ? 
Least ? What relation has the steepness of the graph to the speed of 
the body ? Why ? What are tlie natural limitations on the range of the 
variable ? 

5. In Fig. 13, the beginning of the J.-scale does not appear on the 
graph. Why ? 

6. Rate of increase. In the function of § 21, when < = 2, we have 
A = 1100. Starling with this initial value of t, let f be increased by 1, by 

2, by 3, ••• The corresponding values of A (i.e. the values of A when 
^=2 + 1 = 3, 2 + 2=4, etc.) are respectively 1150, 1200, 1250, •■., and 
the corresponding increases in A over the initial value 1100, are 50, 100, 
150, •••. We see then that for these values the increase in ^ is always 
equal to 50 times the corresponding increase in t.* Show that the same is 
true if we start with a different initial value of f, say t = 3. Prove, in 
general, that starting with any particular value, say t '= ti, of t, and any 
increase in t, say an increase equal to h, that the resulting increase in A 
is equal to 60h; i.e. that the ratio 

increase in A ^q 

corresponding increase in t 

* When a change in the value of the variable produces a certain change in 
the value of the function, these two changes correspond to each other. We 
may then speak of either change as corresponding to the other. 



26 MATHEMATICAL ANALYSIS [I, § 25 

7. From the result of Ex. 6, show that the graph of the function there 
considered is a straight line. 

8. Make an investigation similar to that in Ex. 6 for the function y=x^ 
considered in § 22 ; i.e. calculate the increase in y due to an increase 
in X, under a variety of conditions. For example, let x = 2 initially, and 
calculate the increases in y resulting from increases of 0.5, 1.0, 1.5, 2.0 in 
X. For each case calculate the ratio : 

increase in y 



corresponding increase in x 

Is this ratio constant ? Is the increase in y due to an increase in x of 1.0 
the same when the initial value of x is 3 as it is when the initial value of x 
is 2 ? How is the change in the steepness of the graph related to your 
result ? 

9. A car begins to move and gradually increases its speed in such a 
way that in x sec. it has traveled y = x^ ft. Interpret in this new setting 
the "increase in y due to a certain increase in x," as computed in the 
preceding exercise. Show in particular that the "increase in y" is the 
distance traveled by the car during the interval of time represented by 
the corresponding " increase in x," and that the ratio 

increase in y 

corresponding increase in x 

is the average speed of the car during this interval. Does this suggest 
a method for computing approximately the speed of the car at a given 
instant ? 

10. A certain function y has the value 0, when the variable x is 0, and 
has the value 4, when x = 2. The graph of the function is a straight line. 
Draw the graph and tabulate, from the graph, the values of y when x=l, 
3, 4, 5, C. What is the algebraic relation between y and x ? 

11. The graph of a certain function is a straight line. Draw this 
graph, knowing that y = 0, when x = — 1, and that y = 4, when x = 3. 
Discover the equation connecting y and x. 

26. Statistical Graphs. One of the most generally familiar 
uses of the graph is in connection with the representation of 
statistical data. The figure below represents the enrolment in 
Dartmouth College during the years 1905-1915. The method 
of its construction should be clear without further ex- 
planation. 



I, § 26] RPJPRESENTATION OF FUNCTIONS 



27 



An essential difference between this sort of graph and those 
previously considered must, however, be noted. Strictly speak- 
ing, the graph consists only of the points forming the corners 



-\ U-4-^-f--r-[-f| > H--J -^ H 1 




^ 


1400 --T - " -"f- 


T :--: z-'-'--/ 


:::::::::::::::::: ,,-::„:^^: :::::::: 


„ 1300 J -r ± ±'-:^rk^ - 




^ ::::::::::::::::::::::::::::ii-"::L4::::::::: :::::::::: 


S 1200 ;H; -p^ 1 - - 


CO ^ r**''' 


"^ 1 1 1 1 1 1 1 1 li^44Rfem 1 1 1 1 1 1 ! 1 1 1 mm 1 1 1 1 1 1 1 1 1 nTrrl 


^ 1100 2 "'" X " 


1 ::::>^::::;^:^:._± 


^ ^"^l \y(i[[\\\\\ III 






900 







1905 1906 1907 1908 1909 1910 1911 1912 1913 19li 1915 Years 

Enrolment of Dartmouth College, 1905-1915 
Fig. 16 

of the broken line in the figure. The dates, 1905, 1906, ••• refer 
to the beginning of the college year in September of the years 
given, and the points plotted give the enrolment at the begin- 
ning of each such year. The straight lines joining these points 
are drawn merely for convenience, as an aid to the eye in follow- 
ing the changes in the enrolment from year to year. The points 
of these lines between the end points have no significance. The 
range of the variable here consists of the finite number of dates, 
1905, 1906, •••, 1915; and the function considered is discontin- 
uous. In such a graph interpolation is obviously impossible. 



Questions 

(1) During what periods did the enrolment increase ? decrease ? 

(2) What was the percentage of increase during the 11 years ? 

(3) What was the average rate of increase (in students per year) from 
1905 to 1915? 



28 



MATHEMATICAL ANALYSIS 



[I, § 26 



(4) If the first point of the graph (1905) be joined to the last point 
(1915) by a straight line (see figure), how is the steepness of this line re- 
lated to the average rate of increase ? 



EXERCISES 

1. The maximum temperatures (in degrees Fahrenheit) at Hanover, 
N.H., on successive days from Oct. 1 to Oct. 15, 1914, were respectively as 
follows : 

59.6, 74.8, 79.7, 82.1, 78.9, 66.6, 61.4, 73.7, 82.5, 73.2, 78.9, 66.8, 55.0, 
67.0, 63.5. 

Construct a graph representing these data by a broken line. Is inter- 
polation possible ? Why ? 

2. American shipping statistics give the total iron and steel tonnage 
built in the U.S. for the years 1900-1914 as follows : 



Year 


Tonnage 


Year 


Tonnage 


Year 


Tonnage 


1900 


196,851 


1905 


182,640 


1910 


250,624 


1901 


202,699 


1906 


297,370 


1911 


201,973 


1902 


280,362 


1907 


348,555 


1912 


135,881 


1903 


258,219 


1908 


450,017 


1913 


201,665 


1904 


241,080 


1909 


136,923 


1914 


202,549 



Draw the graph. Is interpolation possible ? Why ? 

27. Summary. As has already been sufficiently indicated, 
the object of our work thus far has been to make clear the con- 
cept of a function. To this end we have considered a variety 
of special functions. Confining ourselves at present to the con- 
ception of what we have had occasion to define. as a single- 
valued function of one variable, we have seen that the essential 
characteristic of such a function may be defined as follows : 

A variable y is said to be a function of another variable x, if 
when a value of x is given, the value of y is determined. 

A variable is a quantity which throughout a given discussion 
assumes a number of different values. The values which a 



I, § 28] REPRESENTATION OF FUNCTIONS 29 

variable may assume constitute the range of the variable in 
question. 

The range of a variable may be limited or not according to 
circumstances. 

We have become acquainted with three methods of repre- 
senting a function : the analytic, the tabular, and the graphic. 

We have made a beginning in the classification of functions : 
single-valued and multiple-valued functions ; continuous and dis- 
continuous functions ; increasing and decreasing functions ; 
functions of one variable and of more than one variable. 

We have had occasion to note some of the questions that may 
arise in the consideration of a function : To determine the 
value of the function when the value of the variable is given ; 
the converse problem, to determine the value (or values) of the 
variable, corresponding to a given value of the function. Both 
of these problems may involve the process of interpolation. 
The maximum or minimum value of a function (and the 
value of the variable for which this maximum or minimum 
occurs) is often of importance. So also is the rate at which 
a function changes its values. This, we have seen, is in- 
timately connected with the steepness of the graph of the 
function. 

28. Algebra as a Tool. The methods to be used in the 
future for the study of functions and their applications group 
themselves naturally under three headings corresponding to 
the methods of representing a function : graphs, analysis, 
tables. 

The first of these we have already considered. It has the 
advantage of presenting the variation of the function vividly 
to the eye ; in this respect it is the superior of either the 
tabular or the analytic method of representation. It lacks 



30 MATHEMATICAL ANALYSIS [I, § 28 

precision, however, since any graph drawn on a piece of paper 
is in the nature of the case an approximation.* 

The analytic representation by means of a formula we have 
touched only very briefly. One of its chief advantages is that 
of the utmost precision and conciseness. This very conciseness, 
however, tends to obscure the properties of the function. The 
tools which enable a sufficiently skillful operator to bring out 
the hidden properties inherent in a formula are comprised in 
what is known as mathematical analysis, of which the processes 
of elementary algebra form the foundation. 

The more important functions have been tabulated. Such 
tables are used primarily to facilitate numerical computations. 
We shall have occasion to use tables frequently. 

The next chapter is devoted to a brief discussion of certain 
algebraic processes and of their relation to the graphic rep- 
resentation already discussed. 

QUESTIONS FOR REVIEW AND DISCUSSION 

1. Give examples from your own experience of quantities that are 
functionally related. In each case, state as many properties of the function 
as you can (continuous or discontinuous, increasing or decreasing, etc.). 

2. State some general laws and discuss the functional relations they 
illustrate. 

3. Would it be desirable to define a function as follows : y is a function 
of x, if y changes its value whenever the value of x changes ? Why ? 

4. Give, from your experience, concrete examples of the use of an 
arithmetic scale. Of an algebraic scale. What are the distinguishing 
characteristics of these two scales ? 

5. Describe the three methods of representing a function and discuss 
the advantages and disadvantages of each. 

6. If the graph of a function y of x is a straight line, and the value of 
the function is known for a; = 4 and for x = 5 (say these values are 20 
and 26, respectively), how can the value of the function for aj = 4.5 be 
calculated (not read from the graph) ? For x = 4.2 ? For a: = 6.7 ? 

*0n the other hand, we can conceive, theoretically, of a graph which is en- 
tirely accurate. 



I, § 28] REPRESENTATION OF FUNCTIONS 



31 



MISCELLANEOUS EXERCISES 

1. The following table gives the pressure of wind in pounds per square 
feet in terms of the velocity of the wind in miles per hour : 



Miles per hour 


5 


10 


15 


20 


30 


40 


50 


60 


70 


80 


Lb. per sq. ft. 


0.1 


0.5 


1.1 


2.0 


4.4 


7.9 


12.3 


17.7 


24.1 


31.5 



Represent the function graphically. Determine approximately the 
velocity which will produce a pressure of 10 lb. per square feet. What 
does the increasing steepness of the curve signify ? 

2. The following table, prepared by the U.S. "Weather Bureau, gives 
the average monthly values of relative humidity at the stations given : 







C3 


< 


5 


>< 


•-5 




6 

-< 




5 

o 


> 

o 


p 


New York 


75 


74 


71 


68 


72 


72 


74 


75 


76 


74 


75 


74 


Chicago . . 


82 


81 


77 


72 


71 


73 


70 


71 


70 


72 


77 


80 


New Orleans . 


79 


80 


77 


75 


73 


77 


78 


79 


77 


74 


79 


79 


San Francisco 


80 


78 


78 


78 


79 


80 


84 


86 


81 


79 


77 


80 



Plot on the same sheet of paper. Is interpolation possible ? Why ? 
3. The following table gives the average weight of men and women for 
various heights : 



Height . 


Weight in Lb. 


Height 


Weight in Lb. 


Men 


Women 


Men 


Women 


5 ft. 

5 ft. 2 in. 
5 ft. 4 in. 
5 ft. 6 in. 


128 
131 
138 
145 


115 
125 
135 
143 


5 ft. 8 in. 

5 ft. 10 in. 

6 ft. 

6 ft. 2 in. 


154 
164 
175 
188 


148 

160 
170 



Represent the two sets of data on the same paper and draw any conclu- 
sions that seem reasonable. Is interpolation possible ? Why ? 

4. The attendance at a base ball park on successive days was as follows : 
1002, 1800, 1875, 1375, 1500, 2750, 3520. Represent these data by points 
in a plane. Is a curve drawn through these points of any significance ? 
Explain your answer. 



32 



MATHEMATICAL ANALYSIS 



[I, § 28 



5. The London Economist gives the following table showing the net 
tonnage of steamships and sailing vessels on the register of Great Britain 
and Ireland from 1840 to 1912 : 



Year 


Steamshii' 


Sailing 

Vessel 


Year 


Steamship 


Sailing 

Vessel 


1840 
1860 
1880 
1900 


87,930 

464,330 

2,723,470 

7,207,610 


2,680,330 
4,204,360 
3,851,040 
2,096,490 


1909 
1910 
1911 
1912 


10,284,810 
10,442,719 
10,717,511 
10,992,073 


1,301,060 
1,112,944 

, 980,997 
902,718 



Represent these data graphically on the same sheet of paper. What 
fact does this graph vividly portray ? 

6. The temperature drop t below 212° at which water will boil at differ- 
ent elevations and the elevation h in feet above sea level are connected by 
the relation h =1"^ + 517 1. Construct a table of values of h ior t = 0, 5, 
10, 15, 20, 25, 30, and draw the graph of 7i as a function of t. At what 
temperature will water boil on Pike's Peak, 14,000 feet above sea level ? 
About how high is it necessary to go in order that water will boil at 200"^ ? 



CHAPTER II 

ALGEBRAIC PRINCIPLES AND THEIR CONNECTION 
WITH GEOMETRY 

29. Numbers and Measurement. We have already had 
occasion to distinguish between two kinds of numbers : 

(a) Numbers each of which represents a magnitude only ; 

(6) Numbers each of which represents a magnitude and one 
of two opposite senses, i.e. the so-called signed numbers. 

It seems desirable at this point to recall the familiar classifi- 
cation of these numbers and the way in which they serve to 
give the measures of magnitudes. We confine ourselves first 
to the numbers of Type (a) above. 

Integers. The first numbers used were the so-called whole 
numbers or integers, 

1, 2, 3, 4, ..., 

which represent the results of counting and answer the ques- 
tion : How many ? They also represent the results of measure- 
ments, when the magnitudes measured are exact multiples of 
the unit. 

The Rational Numbers. When the magnitude measured 
is not an exact multiple of the unit of measure, other num- 
bers called fractions must be used. 

•' A\ 1 1 1 1 *B 

These numbers are intimately asso- c^ . 1 1 ,/> 

ciated with the idea of a ratio. ^' ^^ 

Fig. 17 
Thus, in geometry, two line seg- 
ments AB and CD are called commensurable, if there exists 
a third segment PQ of which each of the other two is an 
D 33 



34 



MATHEMATICAL ANALYSIS 



[II, § 29 



exact multiple (Fig. 17). PQ is then called a common measure 
of AB and CD. If AB is exactly m times PQ and CZ> is 
exactly n times PQ, m and n being integers, we say that the 
ratio of AB to CD is m/n, and we write 

AB^m 
CD n' 

If CD is the unit of length, we have 
the measure ofAB — — • 



A number which can be written as a fraction in which the 
numerator and denominator are both integers is called a 
rational number* 

Such numbers suffice to represent the measure of any magni- 
tude which is commensurable with the unit of measure. 

The Irrational Numbers. If two magnitudes have no 
common measure, they are called incommensurable. Thus we 
know from our study of geometry that 
the diagonal of a square (Fig. 18) is not 
commensurable with one of its sides.f 
Hence, the length of the diagonal of a 
square whose side is 1 unit cannot be 
expressed exactly by any rational num- 
ber. To meet this deficiency the so-called 
irrational numbers, such as the V2, were 
introduced. 

It is beyond the scope of this book to treat irrational num- 
bers fully. But we may note that they serve to express the 

* Observe that according to this definition the rational numbers include 
the integers. The number "zero" is also classed among the rational num- 
bers. See § 30. 

t If AB and AC had a common measure /, such that AB = m x I and 
AC = nX I, where ?n and n are integers, it would follow that n^ = 2in'^ ; but 
this relation cannot hold for any integers in and «. Why ? 




II, § 29] ALGEBRAIC PRINCIPLES 35 

ratio of pairs of incommensurable magnitudes, and, in particular, 
to express the measure of any magnitude which is incommensur- 
able with the unit. 

Moreover, any irrational number may be rejjresented approxi- 
mately by a rational number with a7i error which is as 
small as we please. This follows from the following con- 
siderations. 

It is important to note that the result of any actual direct 
measurement is always a rational number. For example, in 
measuring a distance, we use a foot rule marked into fourths, 
or eighths, or thirty-seconds of" an inch, or else some more 
accurate instrument divided into hundredths or thousandths of 
a unit, and we always observe how many of these divisions are 
contained in the length to be measured. The result is, therefore, 
always a rational number m/n where n represents the number 
of parts into which the unit was divided. Any such actual meas- 
urement is, of course, an approximation. The greater the ac- 
curacy of the measurement (and this accuracy depends among 
other things on the number of divisions of the unit) the closer is 
the approximation. Since we may think of the unit as divided 
into as many divisions as we please, we may conclude that any 
magnitude can be expressed by a rational number to as high a de- 
gree of accuracy as may be desired. Thus, the length of the 
diagonal of a square whose side measures 1 in. is expressed 
approximately (in inches) by the following rational numbers : 
1.4, 1.41, 1.414, 1.4142. These decimals are all rational ap- 
proximations, increasing in accuracy as the number of decimal 
places increases, to the irrational number V2.* 



* Surds, i.e. indicated roots of rational numbers, are not the only irrational 
numbers. The familiar n = 3.14159 ••• is an example of an irrational number 
which is not expressible by means of any combinatiou of radicals affecting 
rational numbers. 



36 MATHEMATICAL ANALYSIS [II, § 30 

30. The Number System of Arithmetic. The (unsigned) 
rational and irrational numbers, together with the number zero 
(which is counted among the rational numbers), constitute the 
number system of arithmetic. 

31. The Nimiber System of Algebra. Corresponding to 
any unsigned number a (except 0) there exist two signed 
numbers + a and — a. The magnitude represented by a 
signed number is called the absolute value of the number, and 
is indicated by placing a vertical line on each side of the 
number. Thus the absolute value of + 5 and of — 5 is 5 ; in 
symbols, |+5|=| — 5|=5. 

The signed numbers are called rational or irrational accord- 
ing as their absolute values are rational or irrational. The 
entire system of positive and negative, rational and irrational, 
numbers and zero * is called the 7'eal number system and any 
number of this system is called a real number. These 
numbers are contained in the so-called number system of 
algebra.^ 



* Note that zero is neither positive nor negative. It has no sign. 

t The number system of algebra contains also the so-called imaginary or 
complex numbers, which will be discussed later. It may be noted that the 
words rational, irrational, real, imaginary, are here used in a technical 
sense. The popular meanings of the terms have no significance. V2 is no 
more " irrational " (i.e. absurd or crazy) than the number 2 ; and the im- 
aginary numbers are just as "real" in the popular use of the term as are the 
(technically) real numbers. Historically, the reason for the use of these 
words is, however, connected with their customary meaning. For, while the 
integers and rational numbers are of great antiquity, the irrational numbers 
were not hitroduced until about the fifteenth century a.d., although incom- 
mensurable ratios were discussed by the ancient Greeks. At that time their 
nature was not thoroughly understood, and it was not unnatural then to 
designate them as irrational. Similar remarks could be made about the 
introduction of the imaginary numbers toward the end of the eighteenth 
century. We may add that what we now call "negative" numbers were in 
the fifteenth century often referred to as " fictitious numbers." 



II, § 32] ALGEBRAIC PRINCIPLES 37 

32. Geometric Representation. Coordinates on a Line. 

It follows from § 29 that the rational and irrational numbers 
are just sufficient to express the length of any line segment. 
Every segment on a line having one extremity at a given 
point or origin can be represented by such a number; 
and every such number will determine a definite one of these 
segments, the unit of measure having been previously chosen. 

This leads at once to the idea of an arithmetic scale, if we 
confine ourselves to the numbers of arithmetic, and to the idea 
of an algebraic scale, if we choose one of the directions on the 
line to be positive, and use the real numbers of algebra to 
represent the (now) directed segments. In the future we shall 
generally confine our discussion to the algebraic case. No 
confusion need arise from regarding an arithmetic scale as the 
positive half of an algebraic scale, nor from regarding the 
numbers of arithmetic as equivalent to the positive numbers 
(and zero) of the real number system.* 

It is often convenient to regard the number x which origi- 
nally represented the length and the direction from to a 



P 

Fig. li) 

point P of the line as representing the point P itself, in which 

case we call x the coordinate of P (Fig. 19). When we have 

chosen a point as origin, selected a unit of length, and 

specified which of the two directions on the line is positive, we 

say that we have established a system of coordinates on the 

line. When this has been done, every point P of the line is 

represented by a number, and every real number represents a 

definite point of the line. 

* For this reason we shall often omit the + sign in writing a positive 
number ; e.g. write simply 5 for + o. The context will always tell whether 
the number in question is signed or not. 



38 



MATHEMATICAL ANALYSIS 



[II, § 33 



33. Coordinates in a Plane. We may now give the precise 
mathematical formulation of the process already used (in con- 
nection with the construction of the graphs of functions) for 
" plotting " points in a plane. The essential features of this 
process are as follows (Fig. 20). We locate arbitrarily in the 









^ 










Secona 


quadrant 


First quadrant 












+ 5 


,Pi 


















^ 


M.O 




t/, -^ 


31, 




X' -5 


; " , J 


X 




quadrant 


^s .5 


■ 


Pi 




Third 


Fourth quadrant 










Y' 











Fig. 20 

plane two algebraic scales, a horizontal one called the x-axiSj 
and a vertical one called the y-axis. These two scales, called 
the axes of reference^ intersect in the zero point of each scale ; 
this point is called the origin. The position of any point P in 
the plane is then completely determined if its distance and 
direction from each of these axes is known. The units on the 
two scales are arbitrary ; they may or may not be equal to 
each other. The distance from either axis must, however, be 
measured in terms of the unit of the other axis, i.e. of the axis 
parallel to which the measurement takes place. Thus, in Fig. 
20, where the units on the axes are different, the point Pj is 
determined by the distance a; = 3 units from the y-axis (meas- 



II, §34] ALGEBRAIC PRINCIPLES 39 

ured in terms of the aj-unit) and the distance y = 2 units from 
the a>axis (measured in terms of the y-\mit). Similarly, the 
points P2, P3, P4 are determined respectively by the directed 
segments OM2 and M2P2, OM^ and M.^P^, OM^ and M^P^ ; the 
numbers representing these directed segments are signed 
numbers, so that the number gives both the magnitude and the 
direction of the segment. In such a system of rectangular 
coordinates in a plane, unless specifically agreed on otherwise, 
the positive direction on the a;-axis is always to the right; on 
the 2/-axis, always upward. 

We see, then, that every point in the plane is determined 
uniquely by a pair of numbers, and, conversely, that every 
pair of (real) numbers determines uniquely a point in the 
plane. The two numbers thus associated with any point in 
the plane are called the coordinates of the point ; the number 
X (giving the distance and direction from the ?/-axi^) is called the 
x-coordinate or the abscissa of the point, the number y (giving 
the distance and direction from the aj-axis) is called the 
y-cobrdinate or the ordinate of the point. Any point P in 
the plane may then be represented by a symbol {x, y), where the 
abscissa of P is written first in the symbol and the ordinate of 
P is written last. Thus we may write (Fig. 20) Pi =(3, 2) 
P, = (- 1, 4), P3 = (- V2, - 31), P, = (?, ?). 

The two axes divide the plane into four regions called 
quadrants, numbered as in the figure. The quadrant in which 
a point lies is completely determined by the signs of the 
coordinates of the point. Thus, the first quadrant is charac- 
terized by coordinates (-{-, +), the second quadrant by 
( — , +), the third by ( — , — ), and the fourth by (+, — ). 

34. Relations between Numbers. If two numbers a and 
b represent two points A and B respectively on an algebraic 
scale, we say that a is less than b (in symbols, a <b), it a is to 



40 MATHEMATICAL ANALYSIS [II, § 34 

the left of b, the scale being horizontal and the positive 
direction being to the right.* The following obvious relations 
are fundamental : 

(1) If a =5t 6, then either a < &, or 6 < a. 

(2) If a < 5 and 6<c, then a<c. 

EXERCISES 

1. Is the date 1916 a signed number ? (Does it represent simply a 
duration of time or does it represent a time after some arbitrary fixed 
time ?) Would it be proper to represent the year 50 a.d. by + 50 and 
the year 50 b.c. by — 50 ? 

2. When we designate the time of day as " two o'clock," is " two" a 
signed number ? 

3. Are the (unsigned) integers used for any other purposes than to 
express the result of counting or measuring ? ( House numbers, catalog 
numbers, •••) 

4. State some theorems of geometry concerning ratios. 

5. Find a rational approximation of VS accurate to within 0.001. 

6. Why is any actual measurement necessarily an approximation ? 

7. Why is it incorrect to define a rational number as one " which does 
not contain radicals ? " 

8. Why should irrational numbers be used at all, if it is possible to 
represent any such number by a rational number to as high a degree of 
approximation as may be desired ? 

9. Explain /rom the definition of ratio why | in. and f^ in. represent 
the same magnitude. Why m/n in. and pm/pn in. represent the same 
magnitude, 

10. Two segments measure | in, and f in., respectively. Show that 
the ratio of the first to the second according to the definition is y^^. (Ob- 
serve that ^ in. is a common measure of the two segments.) 

11. Two segments measure m/n and p/q in. respectively. Prove that 
the ratio of the first to the second is mq/np. (Find a common measure 
of the two segments.) 

12. Given that |a|<|6|, can we conclude that a<6? Why? 
Given that | a | > | 6 1, can we conclude that a > & ? Why ? 

* Likewise, a is greater than b (in symbols, a > b), if ^ is to the right of 
B. Obviously, if a < 6, then 6 > a. 



II, § 35j ALGEBRAIC PRINCIPLES 41 

13. Which is the greater, -3 or -4? —S-lor — tr? 

14. Locate on a line the points whose coordinates are 2, — |, |, — 2, 
6. What is the distance between the last two ? What signed number 
represents the directed segment from the point + 5 to the point — 2 ? 

15. Locate in a plane the points (2, 3), (— 2, 3), (2, — 3), (—2,-3), 
referred to a system of rectangular coordinates, the units on the two axes 
bfiing equal. 

16. If the abscissa of a point is positive and its ordinate is negative, 
in what quadrant is the point ? If abscissa and ordinate are both 
negative ? 

17. If the abscissa of a point in a plane is + 2, where is the point ? 
If the ordinate is zero ? What characterizes the coordinates of a point 
on the y-axis ? On the x-axis ? What are the coordinates of the origin ? 

18. The units on the two scales being equal, what is the distance of 
the point (3, 4) from the origin ? Of the point (— 1, 7) ? Of the point 
(2, - 1) ? Of the point (a, b) ? 

35. The Fundamental Operations. We shall now take up 
briefly the fundamental operations of addition, multiplication, 
subtraction, and division, and develop certain geometric inter- 
pretations and applications connected with these operations, 
which are of importance in what follows. 

Addition. We note first that the operation of addition for 
signed numbers has an essentially different meaning from that 
for unsigned numbers. The addition of two unsigned numbers 
expresses simply the addition of magnitudes. Thus, any two 
magnitudes may be represented geometrically by the lengths 
of two line segments. The segment, whose length represents 
their sum, is obtained by simply placing the segments end to 
end to form a single segment. (Compare the process of graphic 
addition described in Ex. 3, p. 7.) 

A signed number, on the other hand, represents a direction 
as well as a magnitude ; it is represented geometrically by a 
directed segment. Consider two signed numbers a and b. 
They will be represented by two directed segments whose 



42 MATHEMATICAL ANALYSIS [II, § 35 

lengths are | a| and 1 5 1, respectively, and whose directions are the 
same or opposite according as the numbers have the same or 
opposite signs. Figure 21 represents the four possible cases. 
The sum a + 6 is represented by a directed segment which 
expresses the net result of moving in the direction represented 



a 


y 1 


b 






I 


1 


a + b 






1 




a\ 


1 


1 


b 


' 


! 








' 



i — ds: 



a+F o,+b 

Fig. 21 

by a through a distance equal to | « |, and then moving in the 
direction of b through a distance equal to \b\. The segment 
representing a -f 6 is the segment from the initial point of 
these motions to the terminal point. (See Fig. 21.) 

The difference in the meaning of addition in the case of unsigned and 
signed numbers is clearly brought out by considering a simple concrete 
example : Suppose you walk to a place five miles distant and back again . 
The total distance you have walked is 5 + 5 = 10 miles. These are un- 
signed numbers. On the other hand, if you represent the trip out by + 6 
and the trip back by — 5, which is entirely proper, the sum (+5)-f-(— 5), 
which is equal to 0, does not represent the distance walked at all, but does 
represent the net result of your walk measured from your starting point. 
The total distance walked is represented by | + 5 | -f- | — 6 |. 

It should be noted that the absolute value of the sum of 
two numbers is not, in general, equal to the sum of their 
absolute values. In fact all we can say in general on this 
point is that 

(1) k + ?>|<|«| + |^|.* 

The equality sign holds only when a and h have the same sign. 
* The symbol "^ is read " is equal to or less than." 



II, § 35] ALGEBRAIC PRINCIPLES 43 

The geometric interpretations on the algebraic scale of add- 
ing a number x to all the numbers of the scale consists of 
sliding the whole scale to the right or left, according as x is 
positive or negative, through a distance equal to | a;|. Figure 22 
illustrates this operation for the value x = — 2. 

Every number in the upper scale is the result of adding — 2 



-4 -3 -2 -1 +1 +2 +3 



'4 -3-2 -1 + i +2+3 +4 
Fig. 22 

to the number below it in the lower scale. Two important 
consequences follow from this interpretation : 

(1) If a <C b and x is any (real) number, then a -{- x <. b -\- x. 

(2) If a point P ivhose coordinate on a line is x is moved on 
the line through a distance and in a direction giveyi by the number 
h, the coordinate x' of its new position is given by the relation 

(2) x' = X + /i. 

An immediate consequence of the meaning of addition in 
the case of directed segments is as follows. If A, B, C are 
any three points on a line, then 

(3) AB -\-BC = AC, 

This relation holds no matter what the order of the pomts 
on the line may be. In fact it is obvious that to move on a 
line from Ato B and then to move from B to (7 is equivalent to 
moving directly from A to C, no matter how the points are 
situated on the line. As a special case of this relation we 
have 

AB-\-BA = 0, or AB = -BA. 



44 MATHEMATICAL ANALYSIS [II, § 35 

Multiplication. The product ab of two signed numbers 
a and b is defined as follows : 

(1) |a*| = |a|.16|. 

(2) The sign of ab is positive or negative according as the 
signs of a and b are the same or opposite. 

The statement (2) involves the familiar law of signs : 

(+)(+)=(+), (+)(-)=(-)(+)=(-). (-)(-)=(+)• 

Geometrically, multiplication by a positive number x is 
equivalent to a uniform expansion or contraction of the scale 
away from or toward the origin in the ratio | « | : 1, according 
as I a; I is greater than or less than 1. 

This statement will become clear on inspection of the follow- 
ing figure (Fig. 23) which gives the construction for the multi- 





FiG. 23 



plication of every number on the scale by x. In the first figure 
X has been taken equal to + 2, in the second equal to -\- ^. 

The geometric interpretation of multiplication by a negative 
number x consist? of a similar expansion or co7itraction in the 
ratio I ic 1 : 1 combined with a rotation of the whole scale about 
the origin through an angle of 180°. For such a rotation 
will change each positive number into the corresponding nega- 
tive number, and vice versa, which the law of signs requires. 

Here again we may note two consequences of importance : 

1. If a < b and x is any (real) number, ax is less than, equal 
to, or greater than bx, according as x is positive, zero, or negative. 

2. If a scale is uniformly stretched (or contrax^ted), the origin 



II, § 35] ALGEBRAIC PRINCIPLES 45 

remaining fixed, in sugJi a way that the point 1 moves to the point 
whose coordinate is a, then the point whose coordinate is x will 
mx)ve to the point whose coordinate is 
(4) X' = ax. 

Subtraction. To subtract a number b from a number a 
means to find a number x such that x-{-b = a. We then write 
x = a — b. 

Such a number x can always be found. Representing a and 
b by directed segments having the same initial point, the 
meaning of addition tells us at once that 

the segment from the terminal point of < — -- 

b to the terminal point of a represents 
the number x sought. (See Fig. 24.) ^'°- ^ 

This shows, moreover, that to subtract a number b is equivalent 
to adding the number — 5.* 

Division. To divide a number a by a number b means to 
find a number x such that bx = a. We then write x = a/b. 

It is always possible to find such a number x, except when the 
divisor b is zero. For we need merely reverse the construction 

given for multiplication (Fig. 23) as 
indicated in Fig. 25, first drawing 
the line joining b on the original 
scale to the point a on the multi- 
i 6\ plied scale and locating the required 

Fig. 25 point x on the multiplied scale by 

a line through 1 on the original scale, parallel to the line ab. 
In particular, we can always find a number x such tha! 

* It may be of interest to recall here the fact that historically the negative 
numbers were introduced in order to make the operation of subtraction 
always possible (i.e. even in the case when the subtrahend is greater than 
the minuend). But from what has just beeu said it appears that the device 
adopted for rendering the operation of subtraction more useful and convenient 
had the additional effect of making this operation unnecessary. 




46 



MATHEMATICAL ANALYSIS 



[II, § 35 



bx = l,ifb=^ 0. This number 1/6 is called the reciprocal of h. 
Hence, to divide by b (b =^ 0) is equivalent to multiplying by 1/6. 
The Case 6 = 0. This case demands careful attention. Since 
• « = for every number x, it follows that the relation • x = a 
cannot be satisfied by any value of x, unless a is also ; and ivill 
be satisfied by every value of x, if a is 0. Hence, by the definition 
of division, the indicated quotient 

x-^ 

has no meaning whatever when a ^0, and no definite mean- 
ing even when a = 0. Hence, we conclude that division by 
zero, being either impossible or useless, is excluded from the 
legitimate operations of arithmetic and algebra. 

36. The Function a/x. The Symbol oo . Whereas we have 
just seen that division by zero is not a legitimate operation, it 
is highly important for us to note what happens to the fraction 
a/x when x assumes values approachiyig nearer and nearer to 
zero ; as long as x does not equal zero, the indicated division 
is possible. We wish then to consider the /;mcfio?i a/x = y for 
values of x near 0. A table of corresponding values of x and y 
is as follows : 



X 


4 


3 


2 


1 


\ 


\ 


-4 


-3 


-2 


- 1 


-I 


-I 


--: 


ia 


i« 


ha 


a 


2a 


4.a 


-\a 


-i« 


-la 


— a 


-2a 


-4a 



Plotting the points (x, a/x) with reference to two rectangular 
axes we obtain Fig. 26, where we have assumed a to be posi- 
tive and have chosen the unit on the oj-axis to be a times the 
unit on the 2/-axis. 

An inspection of the table and the graph shows us that as x 
decreases in absolute value, a/x increases in absolute value; 



II, § 36] 



ALGEBRAIC PRINCIPLES 



47 



more precisely, by choosing x sufficiently small in absolute 
value, a/x can be made as large in absolute value as we please. 

Further, when x = the expression a/x has no meaning. 
We say the fmicUon is not defined for the value x = 0; or, the 
range of the variable of this function does not include the value 



X 



Fig. 26 



The sentence expressed in black-faced italics above is some- 
times written in a species of shorthand : 



00. 



This looks like an equality involving a division by 0. But 
it does not mean any such thing. The expression a/0 as indi- 
cating a division by has already been pronounced illegiti- 
mate. For this very reason we are at liberty to use the 
symbol to mean something else without danger of confusion. 
We accordingly use it as a short way of expressing the values 
of the variable a/x as x is supposed to approach 0. Similarly, 



48 MATHEMATICAL ANALYSIS [II, § 36 

the symbol oo, read "infinity," does not represent a number at 
all, but a variable which increases loithout limit. The above 
equality is, therefore, an equality between variables, and is 
simply a short way of writing the phrase " as the denomina- 
tor of a fraction, whose numerator is constant and different 
from zero, approaches zero, the value of the fraction increases 
without limit in absolute value." Under these circumstances, 
we also say " the fraction becomes infinite." The phrase 
" equals infinity," which is sometimes heard, is very mislead- 
ing and its use should be strictly avoided. 

Returning to our table and graph, we note also that by 
assigning to x a value sufficiently large in absolute value, the 
value of a/x can be made in absolute value as small as we please, 
but not zero. The shorthand expression of this fact is 

00 

or " as the denominator becomes infinite the fraction ap- 
proaches 0." 

37. The Directed Segment P^Pz- As an application of the 
foregoing principles we will now derive a formula which will 
often be used in the future. Let Pi and P^ be any two points 

on an algebraic scale, and let their 
coordinates be x^ and X2, respectively. 
We desire to find the number repre- 
senting the directed segment P1P2 
in direction and magnitude. By definition x^ = OPi, Xo = OP2 
(Fig. 27). Now, by § 35, Eq. 3, we have 

P1P2 = P,0 4- OP2 = - OPi + OP2 

= -Xi + X2, 

or, finally, 

/^i* 2 ^ •^2 — ^1* 

Thus, if Xi = 2 and X2 = 5, X2 — x^ = -\- 3, and we conclude 



P2 


P, 


Xi Q Xi 




Fig. 27 





II, § 38] ALGEBRAIC PRINCIPLES 49 

that the length of the segment P1P2 is 3 units and that its 
direction is positive (i.e. from left to right in the ordinary 
setting). On the other hand, if cci = 3 and Xg = — 4, we have 
X2— X1 — —I, and we conclude that the length of the segment 
is 7 and its direction is negative (i.e. P2 is to the left of Pj). 

38. Concrete Illustration of the Law of Signs. The law of 

signs, as indeed many of the fundamental laws of algebra, is essentially 
a definition, arbitrary from a logical point of view and dictated largely 
on the grounds of convenience. The following concrete example will 
show how in one instance the conventions adopted in the law of signs 
for multiplication correspond to the concrete facts to be described. 

If a train moves at a constant speed of v miles per hour, then in t 
hours it will travel a distance = ^?i miles. Here v, t, s are unsigned num- 
bers. Now, let us change the formulation somewhat, so as to introduce 
the direction. At a given instant let the train be at a certain station 0; 
let us count time from this instant (t = 0) so that any positive t desig- 
nates an instant a certain number of hours after the instant t = 0, 
and a negative t designates an instant a certain number of hours before 
t = 0. Further, let the position of the train be determined by the signed 
number s representing the distance and the ^^ 

direction of the train from 0, s being positive [ ^ ^ 

if the train is to the right of (Fig. 28). |=o *"■*" 

Finally, let the speed and the direction in -p^^ 28 

which the train is moving be given by the 

signed number v, v being positive if the train is moving to the right 
(u = — 30, for example, would mean that the train is moving to the left 
at the rate of 30 miles per hour). 

Now consider the four cases : (1) v and t both positive ; (2) v positive 
and t negative ; (3) v negative and t positive ; (4) v and t both negative. 
Verify that the law of signs in the relation s = vt gives the sign to s 
for which the actual position of the train in each case calls. [For 
example : (1) If v and t are both positive, s = vt will be positive, 
which is as it should be. For if the train is moving to the right, then 
a certain number of hours after t = 0, when the train was at s = 0, it 
will be a certain number of miles to the right of 0. (2) If v is positive 
and t negative, s = vt is negative. This also is correct. For a train 
moving to the right and arriving at O when t = 0, was to the left of 
at any time before t = 0. Etc. ] 



50 MATHEMATICAL ANALYSIS [II, § 38 

EXERCISES 

1. Under what conditions is|a + &| = \a\ +\b\? 

2. Prove that if A, B, C, D, •••, i, M are any points on a line (in 
any order) then AB + BC + CD + •" + LM = AM. 

3. Graphic Addition. Given the directed segments, a, 6, c, d, e on 
parallel lines (or on the same line), their sum a+6+c+cZ + e may be 
found graphically as follows: On the straight edge of a piece of paper 
mark a point ; lay the strip along the segment «, the point coincid- 

^ - ing with the initial point of a ; mark the ter- 

'' *■ minal point of a on the paper. Then slide the 
paper parallel to itself so as to make it lie 
along h and bring the mark just made into 
coincidence with the initial point of h ; mark 
the end-point of h. Then proceed similarly 
for the segments c, d, e. The directed seg- 
ment from to the final mark will then represent the sum sought. Why ? 

4. Draw directed segments representing the numbers — 3, + 5, + 2, 

— 6, and find their sum graphically. 

5. Show how to construct a directed segment representing the prod- 
uct of the numbers represented by segments a and h. 

[Hint. Use the adjoined figure to determine the 
magnitude of the product ; then determine the direc- 
tion. Observe that for the construction of a product 
we need to know the length of the unit segment, which 
was not necessary for a sum.] 

6. Show how to construct a segment representing a/h. 

7. Determine the numbers representing the directed segments from 
the first point of each of the following pairs of points to the second: -f- 8 
and + 6, + 8 and -6,-2 and - 4, — ^ and + |, + 1.4 and - 2.1, — | 
and - I, + -V- and + 3.14. 

8. By computing the numbers representing the segments, verify the 
relation AB + BC = AC, when the coordinates of A, B, C are, respec- 
tively : 

(a) 2, 3, 4 ; (6) 2, - 3, 4; (c) - 2, 3, - 4 ; (d) -2,-3, 4. 

9. Find the coordinate of the mid-point of the segment joining the 
points whose coordinates on a line are 4 and 8 ; — 3 and 5; — 2 and 

— 5 ; Xi and X2. 




II, § 40] ALGEBRAIC PRINCIPLES 51 

39. Insight and Technique. Most of our activities involve 
two more or less distinct aspects : insight and technique. On 
the one hand, we need to understand the nature of the thing 
we are trying to do, on the other we need skill in doing it. 
Theory and practice, planning and carrying out the plans, etc., 
are other ways of pointing the same distinction. 

In your previous study of arithmetic and algebra the major 
emphasis was on the side of technique. You learned at that 
time how to carry out certain manipulations with numbers ; and 
you gained more or less skill in using the processes. In the 
present course, the emphasis is to be placed on the side of in- 
sight, understanding, appreciation ; the technique of algebra is 
to be used merely as a tool, not as an end in itself.* 

40. Definitions. We propose now to recall very briefly a 
few of the more important conceptions and processes of 
algebraic technique. We shall begin with the definitions of a 
few terms. 

When two or more numbers are added to form a surriy each 
of the numbers is called a term of the sum. 

When two or more numbers are multiplied to form a 2)roduct 
each of the numbers is called a factor of the product. 

Any combination of figures, letters, and other symbols, 
which represents a number, is called an expression. If the 
equality sign (=) is placed between two expressions, the result 
is called an equality, and the two expressions are called the 
members or the sides of the equality. 

An equality states that the two expressions represent the 
same number. -f 

* However, we must maintain a certain amount of proficiency in the use of 
algebraic processes. Hence " drill exercises " will not be wholly lacking in 
what follows. 

t Such a statement may or may not be a true statement. See § 47. 



52 MATHEMATICAL ANALYSIS [II, § 40 

Thus, suppose a, b, c, d, p, x, y represent numbers. Then 

a — hx -{- 1 cdy = a (12 y"^ — p) 

is an equality. The left-hand member is a sum of three terms; 
the right-hand member consists of only one term, which is 
a product of two factors. The second term of the left-hand side 
is a product of two factors, while the second factor of the right- 
hand side is a sum of two terms. 

41. General Laws of Addition and Multiplication. The 

following general laws we take for granted : 
I. Concerning Addition : 

1. Ariy tivo numbers may be added and their sum is a definite 
member. 

2. The terms of any sum may be rearranged and grouped in 
any way without changing the sum. 

Thus, if a, &, c, p, q represent any numbers whatever, we 
have, for example, a -{-{b + c -^p) -\- q ={b -{- q) -\- {c -\- a) -\- p. 

II. Concerning Multiplication : 

1. Any two numbers may be nudtiplied and their product is a 
definite number. 

2. The factors of any product may be rearraiujed and grouped 
in any way without changing the product. 

Thus, if a, 6, c, x, y represent any numbers whatever, we 
have, for example, {abc){axy)= a^bcxy ={yx){cba'^). 

III. The Distributive Law : To multiply any sum by any 
number m, we may multiply each term of the sum by m and add 
the resulting j^foducts. 

Thus, (a -{- b -\- cd -\- — -\- x)m = am -}- bm -f cdm + — + xm. 

lY. The Law of Factoring : If every term of a sum con- 
tains the same number m as a factor , the sum contains m as a 
factor. 

Thus am -\-bm -^ cdm -f ••• -f xm = m (a -f- & + cd + — -f x). 



11. § 43] 



ALGEBRAIC PRINCIPLES 



53 



Observe that IV is obtained from III by simply interchang- 
ing the sides of the equality. 

42. Raising to Powers. Integral Exponents. We recall 
also at this point the meaning and use of integral exponents. 
The symbol cc", where x represents any number and n is any 
positive integer, is an abbreviation for the product of n factors 
each equal to x, i.e. 

x"" = x- X'X-" to n factors. 

From this definition and Principle II (§ 41) it follows at 

once that 

/pm^n =(x'X' X '" to m factors) (x • x 'X ••• to n factors) 

= X ' X • X "• to m + n factors, 
and therefore 

Ya x»^x^ = x^^'\ 

Similarly 
V6 



— = x^^'^, if m > n, 

Xn 



Xn ;t" 



if n > m. 



(xmy = x"" • x"" ' x"^ •" to n factors 



r^m+m+m+ ■•■ ton term" 



and 

Also 

and therefore 
VI 

Also 
Vila 

VII6 



43. Axioms. Closely connected with Principles I, 1 and 
II, 1 are the familiar axioms 

VIII a If a = h and c = d, then a -^ c = b -^ d. 

VIII b Ifa = b and g = d, then ac = bd. 



(^m)n 


= x**'^^. 




= a»b*K 



64 MATHEMATICAL ANALYSIS [II, § 43 

EXERCISES 

1. Distinguish between insight and technique in the various professions 
(surgery, dentistry, engineering, etc.). 

2. Complete the following propositions : 
(a) The sum of any two integers is • • • 
(6) The product of any two integers is • • • 

3. What is the familiar expression in words for Principle VIII ? 

4. Find the results of the following indicated operations : 

(1) a;i0a;i2. (6) x^^-^x^ (11) (av^y. 

(2) a^a^. 0) a^-^a^^. (12) (- cM»y. 

(3) ft^fes. (8) -^. (13) (-0*. 

a" 

(4) y2ny3«. ^g-j ^^4^8. (14) (r^S-)". 

(5) X'»-%2. (10) (C2)6. ' (15) (-X2)». 

5. Multiply x^* + x^y^ + i/2& by x2» — x'^y^ + y"^. 

6. Divide x^" + y^n by x" + ?/". 

7. Perform the following operations : 

(1)25.24= (2)25.44= (3)32.23= (4)7i5^7i3 = 

44. Discussion of Principles. In the preceding article 
Principles V-VII were derived from I and II, while IV is a 
consequence of III. We might now ask : " How do we know 
that Principles I, II, and III are true for all numbers ? " 

On these three principles the whole subject of algebraic 
technique rests. They are so simple that they may appear at 
first sight to be trivial. As a matter of fact their truth is by 
no means obvious ; our unquestioned belief in them is the re- 
sult of experience in using numbers. Were we to attempt a 
general proof, we should find it a long and difficult process 
which is out of place in an introductory course. Hence we 
simply take them for granted. 

A little reflection will show that these principles are not obvious. Take 
for example the fact implied by Principle II, 2 : a times b is equal to h 
times a ; and let us suppose that a and b are positive integers. Now, 
2x3 means 3 + 3 and 3x2 means 2 + 2 + 2. By addition we observe 



II, § 44] 



ALGEBRAIC PRINCIPLES 



55 



that the result is in both cases 6. But that simply verifies the general 
law when a = 2 and 6 = 3. We can thus verify the law in question for 
any two special integral values of a and h. Not only would this be ex- 
tremely laborious for large values ; it would still be only a verification for 
a special case ; it would not be a general proof. Moreover, we have con- 
fined ourselves to the simplest of all numbers, the positive integers ; while 
II, 2 asserts among other things that ah = 6a, no matter what numbers 
a and 6 represent (rational or irrational, real or imaginary) . As has been 
indicated in the preceding paragraph, we are not concerned in this course 
with proving these principles. Of great interest to us, however, are the 
relations existing between numbers and geometry. Accordingly we have 
suggested in the exercises below some geometrical interpretations of these 
principles. which furnish intuitive proofs of certain restricted cases. 



EXERCISES 

1. An intuitive proof that ah = ha, in case a and 6 are positive inte- 
gers : Let the integer a be represented by a group of a equal squares 
placed side by side so as to form a row (see the figure, where a = 8). 



The product 6 • a is then represented by the number of squares in 6 such 
rows. Show that, if these rows be placed under each other (as in the 
figure, where 6 = 6), it is seen that the total number of squares is also 
equal to the number of squares in a columns each containing 6 squares. 
Observe that while the figure is drawn for special values of a and 6, the 
argument is general. 

2. From a consideration of the adjacent figure 
give an intuitive proof that 5 • (3 • 4) = 3 • (5 • 4) . 
Then by using the fact that ah = ha show that 
(3 • 4) • 5 = 3 • (4 . 5). Can this argument be made 
general to show (a • 6) • c = a • (6 • c), when a, 6, c are positive integers ? 



4 


4 


A 


4 


4 


4 


4 


4 


4 


4 


4 


4 


.4 


4 


4 



66 



MATHEMATICAL ANALYSIS 



[II, § 44 



3. From the adjacent figure, show how to use the idea of a rectangu- 
lar pile of blocks to prove that (a • 6) • c = a • (& • c), when a, &, c are 
positive integers. 





^--^^4^-^ 


7^ 


^ 


•^ 


j^'O^O^O' >' 














^ 


















u^ 



4. Assuming that the area of a rectangle is equal to the product of its 
base by its altitude, show that ab — ba, when a, b are any positive real 
numbers. 



5. By considering the adjacent figure, interpret 



a b 



geometrically the relation {^a ■\- b)c := ac -{■ be. 



6. Interpret geometrically the equalities 

(a) (a + &)2 = a2 + 2 a6 + &2. 

(6) (a + 6) (c + (?)=: ac + he ^- ad ^- bd. 



d 

€ 



7. Derive the equalities in Ex. 6 from Principles I-V. 

[For 6 (6), we first consider c + d as a single number. Ill then gives 
(a + 6)(c + d) = a(c + d) + 6 (c + d). Applying II and III to each of 
the terms of the right-hand member of this equality, we get the desired 
result. ] 

8. Show in detail how the carrying out of the product (ax+6)(cx+d) 
involves Principles I-V. 

9. Show how these principles apply in the addition of 2 x^ + 7, 3 x + 2, 

and 4 x2 + X -}- 3. 



II, § 45] ALGEBRAIC PRINCIPLES 57 

45. Review of Algebraic Technique. We propose now to 
take up a few of the most elementary portions of the tech- 
nique of algebra. These are all that will be needed in the 
immediate future. Other topics relating to technique will 
be recalled when they are needed. 

The technique of algebra is concerned altogether with the 
transforming of expressions into other equivalent expressions 
which serve better the purpose in hand. The principal pro- 
cesses used are the following : 

(a) Performing indicated operations and collecting terms. For ex^^ 
ample, collect the terms in x, ?/, and z in the following : --^ 

2x4-72/ -3^; + y -\x—^y ^bz^ 3x. T"..''/' 

The result \%x -\-1z. This involves Principles I and IV. 
Perform the indicated operation and collect terms in 

(x2_3a;-|-4)(x-2). 

The result is a;-^ — 5 a;^ + 10 a: — 8. This involves Principles I-V. 

(b) The use of special products. The following equalities should be 

memorized : 

(1) (a-}-b)(a-6)=a2-62. 

(2) (a + b)2 = a^ + 2a& + &2. 

(3) (a-&)2 = a2-2ab + 62. 

(4) (AC + a)(x -h b)= x2 + (a + 6) X + ab. 

(c) Factoring. The following cases may be specially mentioned : 
i. I%e difference of two squares. Use special product (1). Thus 

^9x^ - iy^ ={7 x« + 2y)(7 x^ - 2y). 
ii. Trinomials of the form x^ + px + q. 

Try to find two numbers whose sum is p and whose product is q, in 
accordance with special product (4) . Thus to factor 

x2 _ 6 X - 27 

we notice that 3 and — 9 are two numbers which satisfy the requirements. 
Hence, 

a;2_6x-27 = (x+ 3)(x-9). 

Again, to factor x2 — 10 x + 25, we notice that — 5 and — 5 are two 
numbers satisfying the required conditions. Hence 

x2 _ lOx + 25 = (x - 5) (x - 5) = (X - 5)2. 



58 MATHEMATICAL ANALYSIS [II, § 45 



EXERCISES 
1. Perform the following operations : 



(a) (a; -4) (2 a: + 3). 

(&) (x + a)(x + b)(x + c)' 

(c) (x + h)^-x^. 


(d) (m + w) n — (m — n)m. 

(e) (a + &)2-(a-6)2. 
(/) (a + 6)3 -(a -6)3. 


2. Factor: 




(a) x^-W. 

(&) (2a-by-9(x-iy^. 

(c) ax — bx + ay — by. 

(d) x*- 6x^ + 9. 


(e) x2 + 6x + 6. 

(;i) 25x* + 10a;3 + x^. 


3. Factor: 




(a) 4a2_5« + i. 
(6) a2 + 2a6 + 62_a;2. 
(c) a9-64a3_a6 + 64. 
((^) 1 + x2 + xK 


(e) x3 -3a;2 + 2a;. 
(/;2 + 7x-15x2. 
(Sr) x8 + 1. 
(/i) a:4?/2 _ 17 x'^y - 110. 



46. Operations with Fractions. These depend chiefly on 
the simple principle that the numerator and the denominator 
of a fraction may be multiplied by the same number (not zero) 
without changing the value of the fraction, and the reverse of 
this principle, viz. that any common factor (not zero) of the 
numerator and the denominator of a fraction may be removed 
without changing the value of the fraction.* 

By means of this rule any two or more fractions can be re- 
duced to the same denominator. The rules for adding and 
multiplying fractional expressions are stated symbolically as 
follows : 



a 
b 


+r 


b 


c 


a 
b 


xr 


ac 
'bd 






a 
b~ 


na 
''nb 





* This principle is a direct consequence of the definition of division. Can 
you explain it ? 



II, § 46] ALGEBRAIC PRINCIPLES 59 



IBR^ 


iK 


: PRINCIPLES 




a 
b 


+r 


ad-\- be 
bd 


a 
b_ 


a 
"b 


d 

• — — 

c 


ad 

= — . 

be 


d 








a 

b 


= - 


— a 
b 


a 
-b 




a 


+ &_ 


-a-b 



Also 
and 



c c 

The following exercises will furnish applications of these 
principles. 

EXERCISES 
1. Express as a single fraction : 

^^^ M^ + ^T^ + ^T&' ^ ^ c^y a'^ b^^ 

^ X y z x + y -\-z \n vj\m uj 

(c) L^l + L. (^) 1 + 1_§. , 

St tr rs ^ 2 X y 

Simplify the following expressions, assuming that no canceled factors 
have the value zero. 



a-\-b 



2. 


X - 


-— + 
- a 


{X- 


-aY 


\x 


-ay 


3. 




a - 


- c 






h + c 



(a-b){c-b) {a-c)(a-b) {b-c){c-a) 



Ans. 2« 



(a_6)(c-6) 

«J=L^ _ 1±A I (62 _ an Ans. 4 ab. 

a + b a-&P ^ 

g x^ -y* ^ g-i^ ab + b^ 

' \a^ b^\ ' \a^ b^ 



60 MATHEMATICAL ANALYSIS [II, § 46 



be ac ab a) ( i _ ^ c 

V a + & + c 

\a^-^b^)\ b I \b a) 

^2 ^. 7y2 q;2 _ ft2 



) 



10. 



11. 



aP- - b-^ 


a' + b-^ 


a + b 
a-b 

1 


a-b 
a + b 



^_^x^ 2\x-y x-\-y) 



1+-^+ . "".J M- 



xy + 4 y'^ x^ + 4:xy y(x + 2 y) 



Aiis. a 



111 c U (a + 6)^-c° l 
I a + 6|l (a + 6/ j 

i + l + i 

13. 2_L_2. 

£ + « + * 
a & c 

14 a^ + &'^ ^ ■ ?) \ g^-fe^ 
a^ + rt6 + ^n a- bl a^-ab + b^' 

X^ + -J^\ {X'^ + y^} 
X'^ 1/2 '- 

15. -^ ^-^ Ans. X* - a;2y2 + y*, 

^ I y 
x-hy x — y 

16. If a, ft, and « are positive, which is the greater, 

or 

b (b-{-x) 

Distinguish two cases a>- b^ a <.b. 

17. If ad < eft, then is it true for all values of the letters involved that 
a/b < c/d ? Why ? 



II, §48] ALGEBRAIC PRINCIPLES 61 

47. Identities and Equations. We must recall here a vital 
distinction between two kinds of equalities. An equality 
which is true for all values of the letters (or other symbols) 
involved, for which both members of the equality have a mean- 
ing, is called an unconditional equality or an identity. An 
equality which may be true for certain values of the letters 
involved, but is not true for all, is called a conditional equality 
or an equation. For example, the equality a'^ — b^=(a— b) (a + b) 
is an identity since the two members of the equality represent 
the same number for all values of a and b. Also the equality 

^3^ =^ + ^ 

is an identity, even though it becomes meaningless when a = b. 
Why ? On the other hand 2^ — 8 = is an equation since 
it is true only for x = 4. Va? + 1 = — 1 is also an equation, but 
it is not true for any value of x. Why ? * 

To solve an equation is to find the values of the letters for 
which it is true. Thus in the first example above, a; = 4 is the 
solution or root of the equation 2 a? — 8 = 0. The second equa- 
tion above has no root. 

48. The Relation AB = 0. In the solution of equations 
the following principle is of frequent application. If a prod- 
uct of expressions each representing a number is zero, we may 
conclude that some one of the factors is zero. In the simplest 
case this means that if A and B represent expressions and if 
A • B = Oy we conclude that either ^1 = or J5 = O.f 

* By Va is meant the positive square root of a. 

t We must, of course, be careful to assure ourselves that each of the expres- 
sions involved represents a number for the values under consideration. Thus 
we cannot conclude from the relation x • (l/x) = 0, that either a = or 1/x = 0, 
for when a; = 0, l/x is meaningless. In fact the given relation is impossible; 
the equality is not true for any value of x. 



62 MATHEMATICAL ANALYSIS [II, § 48 

We may apply this principle to show the absurdity of some 
mistakes that are often made by the careless student. For 
example, a favorite mistake is to " cancel " the x in the expres- 
sion 

a-\-x 

This would be justified if the equality 

a\ a + x ^a 

^^ b+x b 

were an identity. If we clear this equality of fractions by 
multiplying both members by (b + x) b, we obtain 

ba '\- bx = ab -{- ax, 
or 

bx = ax; 
or, finally, 

(b-a)x = 0. 

Hence we conclude that equality (4) cannot be true, unless 
either 6 = a, or ic = 0. The " canceling operation " mentioned 
above is therefore unjustified. 

EXERCISES 

1. Treat similarly the following equalities to determine under what 
conditions they are true. Each one is related to an error that is some- 
times made. 



(a) Is Va2 + 6'-^ = a + 6 ? (Square both sides.) 

^ ^ b d b-\-d 

^ ^ 2c + d c + d 
(d) Is (x + 2/)2 = x2 + y2 ? 
(e) Given x^ = 2x. Are we justified in concluding that x = 2 ? 



II, § 48] ALGEBRAIC PRINCIPLES 63 

2. In each of the following equalities, assuming that the letters repre- 
sent real numbers, determine which are identities and which are equations. 
Among the latter, distinguish those that are not true for any (real) values 
of the letters involved ; and for the others determine in their simplest 
form the conditions which they imply on the letters involved. 

(a) a;*-y4=(x + 2/)(x-y)(x2+|/2). 
(6) x'^-Sx + 2 = 0. 

(c) x+- = 0. 

X 

(d) ac — he + ad — bd = 0. 

3. Find and discuss the error in the following reasoning : 

Let X = 2. Then a;2 = 2 x, and x^ — 4 = 2x — 4. This is equivalent to 

(a;+2)(x-2) =2(:b-2). 

Dividing both sides by x — 2, we get 

cc + 2 = 2. 
But X = 2 ; hence 

2 + 2 = 2 
or 

4 = 2. 

4. Find and discuss the error in the following reasoning : Let a and b 
represent two numbers. Then 

a2 _ 2 a6 + 62 =6-2 _ 2 a& + a% 
or 

(a-.6)2=(6_a)2, 
or 

a — b = b— a; 
hence 

a = 6. 



PART II. ELEMENTARY FUNCTIONS 

CHAPTER III 
THE LINEAR FUNCTION. THE STRAIGHT LINE 

49. A Linear Function. Distance traversed at uniform speed. 

Example. A railroad train starts 10 miles east of Buffalo 
and travels east at the rate of 30 miles per hour. How far 
from Buffalo is the train at the end of x hours ? 

In oc hours the train travels 30 x miles. If its distance from 
Buffalo is denoted by y, we have 2/ = 30 a; + 10. Pairs of values 
of X and y obtained from this equation are shown in the fol- 
lowing table. 



X 





1 


2 


3 


4 


etc. 


y 


10 


40 


70 


100 


130 


etc. 



Geometric Eepresentation. Let us plot as points in a 
plane these corresponding values of x and y. We then obtain 
the first of the following figures (Fig. 29). It will be noticed 



Y 


Y 


Y 


































































































I ' 


• 3 4X i i 


^ ^X 12 3 4 



Fig. 29 
64 



Ill, §49] 



LINEAR FUNCTIONS 



65 



that the five points appear to lie on a straight line. We have, 
for intermediate values of x, the values of y shown in the fol- 
lowing table. 



X 


i 


^ 


■ 2i 


H 


y 


. 25 


55 


85 


115 



Plotting these points we obtain the second of the above figures, 
in which the nine points appear to lie on a straight line. Let 
us calculate the value of y for some more intermediate values 
of X thus : 



X 


JL 
4 


f 


i 


i 


y 


m 


?>2i 


47i 


62^ 



In the third figure we see that these new points still appear to 
lie on the same straight line. 

These considerations suggest that if we could calculate the 
values of y corresponding to all the values of x between x = 
and a? = 4, the points whose coordinates 
are {x, y) would all lie on a straight 
line joining the points (0, 10) and 
(4, 130), and would constitute the 
whole of this line-segment. A proof 
that this is the case is as follows : In 
Fig. 30 we have drawn the straight 
line joining the points A (0, 10) and 
B (4, 130). Let (x^, y^)(x, > 0) be any 

pair of corresponding values of x and y for the function 
2/ = 30 ic + 10 ; we then have 
(1) 2/1 = 30 0^1+10. 





. ,,, 




120 -^ 


- J- «! 




i^ 


80 - Al 




60--'- :::^ :n 


^'^ Jz t^ 


40 - ^ ^ 1 ^ 


/^ II 


20 ^ til- ' 


A 4 - yj, ^ 


^ i-iS-i-^-SElL 


1 2 M^ * 



Fig. 30 



66 MATHEMATICAL ANALYSIS [III, § 49 

We now wish to prove that the point P (a^i, y{) is on this 
line AB.* To do this we construct the triangles APQ and 
ABC by drawing lines through A, P, and B parallel to the 
axes. If P is on the line JLB, then these triangles are similar, 
and if P is not on the line AB, then the triangles are not similar. 
Why? If the triangles are similar, QP/CB = AQ/AC; and 
conversely, if QP/CB = AQ/AC, the two triangles are similar. 
Expressed in terms of the coordinates of A, By and P, this pro- 
portion becomes (see figure) 

120 4 ' 

or ^\i~'' 

3/1-10 -120^3^ 

Xi 4 

But from (1) we have y^ — 10 = 30 Xi and hence 

^-Zi? = 30. 

Xi 

This proves that every point whose coordinates {x^, y-^ satisfy 
the relation 2/ = 30 a; + 10 is on the straight line AB. 

Conversely, every point on the straight line AB has coordinates 
{xij yi) which satisfy the relation y = SO x -\- 10. 

For, from the figure, we have 

yi-10 ^120 
Xi 4 

whence 

yi=S0x^-\-10. 

* Extended beyond B, if Xi > 4. 

** Observe that QP and CB are measured in different units from AQ and 
AC. But the ratio of two line-segments is independent of the unit in which 
they are measured. 



Ill, § 49] LINEAR FUNCTIONS 67 

The straight line AB (extended indefinitely beyond B) then 
gives a complete representation of the function 30 a; + 10, at 
least for positive values of x (negative values of x have no 
meaning in this problem). Every pair of corresponding values 
of X and y gives rise to a point on AB, and eveiy point of AB 
has coordinates which are corresponding values of x and y. By 
virtue of this fact the line xiB is called the graph of the function 
30 a; 4- 10, or the locus of the equation y = 30 x -{- 10 referred to 
rectangular coordinates ; Avhereas the equation ?/ = 30 a; + 10 is 
called the equation of the line AB. 

Uses of the Graph. The graph just discussed exhibits 
vividly to the eye several properties of the function 30 a; 4- 10. 

(1) The function steadily increases as x increases. This 
corresponds to the fact that the longer the train moves east- 
ward, the greater is its distance from Buffalo. 

(2) Corresponding to every positive value of a;, there is a 
unique value of y. From the graph find y when x is 4. 

(3) Corresponding to every positive value of y (greater than 
10) there is a unique value of x. What is the value of x when 
.vis 160? 

(4) The last consideration means that x is also a function of 

y. Explicitly we have 

y = 30 a; + 10, 
whence 

2/ - 10 = 30 a; 
and 

x=.y^^. 

30 

•It is left as an exercise to draw the graph of the function 

y-10 
^='"^0~ 

by assigning values to y and computing the corresponding 
values of x. Compare the result with the graph in Fig. 30. 



68 MATHEMATICAL ANALYSIS [III, § 49 

Rate of Change of a Functiox. Before leaving this special 
case to consider a more general problem, we shall use it to illus- 
trate a very important conception connected with a function. 
We have noted that when a; = 0, y = 10. Starting from this 
initial value, as x increases from the value 0, the value of the 
function, i.e. _?/, changes (in this case increases). It is often of 
the greatest importance to know how the increase in the func- 
tion y is related to the increase in x. As x increases from 
to 1, y increases from 10 to 40 ; i.e. a change in x of one unit 
produces a change in ?/ of 40 — 10 or 30 units. The relative 
change is then -^^2^, or 30. As x increases from x = to x = 2, y 
changes from y z= 10 to y= 70, or by 60 units, and the relative 
change is again 30. 

Let us see what the situation is in general. Let Xi be any 
particular value of x and yi the corresponding value of y ; then 
suppose that Xo is any other (subsequent) value of x and ?/2 the 
corresponding value of ?/. The change in x is evidently X2 — Xi 
and the corresponding change in y is y2 — 2/1- We seek the 
value of the ratio 

X^-Xi 

We have from the data of the problem 

2/2 = 30 X2 + 10 
and 

y^ = 30 a^i -h 10. 
Subtracting we get 

2/2-2/1 = 30(^2 -aJi) 
and hence 

^2-3/1^30, 
0^2 — a?! 

We see then that the ratio of a change in the function 30 a; +10 
to the corresponding change in x is constant and is equal to 



Ill, § 50] 



LINEAR FUNCTIONS 



69 



the speed of the train. We shall see presently that in any 
function of the first degree in x, the ratio of a change in the 
function to the corresponding change 
in X is constant. 

Geometrically this result expresses 
the familiar proportionality of homol- 
ogous sides of similar triangles. By 
reference to Fig. 31 we may readily 
verify that the terms 2/2 — 2/i ^^^^ ^2 — ^1 
represent the vertical and the hori- 
zontal sides of a right triangle whose 

hypotenuse is on the line AB. The fact that the ratio 
(2/2 — 2/i)/(^2 — ^1) is constant, i.e. always equal to 30, simply 
corresponds to the obvious fact that any two such triangles, 
no matter at what place they are drawn, or how long their 
sides are taken, are similar. 

















\ l^' 




IV 


inn 


~ A t" 




^^ V 




^-- i 




yfr T^ 


eo ~" ,> 1:1 1 1 \\r\ 1 


bU ^^ 


I ISI 


40 -P -.^ 




4^ : 




20-i'^V Xn- 


= •^1--- + -- 




+ 


^ 2 


3 i 



Fig. 31 



50. Change Ratio. The ratio 

X2 — X1 

is called the change ratio (or sometimes the difference ratio) of 
the function. The difference X2 — x^ is often denoted by Ace, and 
the corresponding difference y^ — 2/1 by Ai/.* The change 
ratio may then be written A?//Aiw. Explicitly, by definition, 
we have the following equalities : 

change ratio = ^ = fclli^ = change in y . 

Ajt X2 —Xi corresponding change m x 

The preceding considerations suggest the theorem : 

* A is a Greek capital letter corresponding to our D and called delta ; it is 
used because d is the initial of the word ** difference." " Ax," is then merely 
an abbreviation for "difference of the x's " or '* change in z " and " Ay " for 
" difference of the y's " or " change in y." 



70 MATHEMATICAL ANALYSIS [III, § 50 

If the change ratio of a function is constant, the graph of the 
function is a straight line; and conversely. 

The truth of this theorem is already sufficiently indicated in 
case 2/2 — Vi ^^d ^2 — ^1 3.re both positive numbers. In formu- 
lating a general proof we must keep in mind that y^ — 2/1 ^^d 
X2 — Xi may be either positive or negative and that these dif- 
ferences represent directed segments. The proof of the theorem 
in general will appear presently. 

EXERCISES 

1. Discuss fully the graph of the function y = 2 a; + 3. Prove that the 
graph is a straight line. Express x as a function of y. Find the change 
ratio and show that it is constant. 

2. Proceed as in Ex. 1 for each of the functions : 
(a) 5x4-2, {h)x+\2, (c) 3.2a: + 8.4. 

3. Prove that the change ratio for the function y = mx + 6 is m. 

4. A steamer 150 miles east of Toledo starts to travel west at a uniform 
rate of 15 miles per hour. Express its distance y east of Toledo at the end 
of X hours. Draw the graph of the function and prove that it is a straight 
line. Does the distance y increase as x increases ? Calculate the change 
ratio and show that It is constant. What is the significance of the 
negative sign ? At what time is the steamer 10 miles east of Toledo ? 
When does it reach Toledo ? How are the last two results shown in the 
graph ? What is the significance of the graph that extends below the x-axis ? 

5. Give examples, drawn from your experience, of functions which 
(a) increase as the variable increases ; 

(6) decrease as the variable decreases. 

6. Consider the function y ■= x^. Calculate the corresponding values 
of y when a; = 0, 1, 2, 3, 4, 5. Plot the corresponding points and observe 
that they are not on a straight line. Calculate the change ratio of this 
function for x = and Ax = 1, 2, 3, and observe that it is not constant. 

7. The cost of printing certain circulars is computed according to the 
following rule. The cost for the first one hundred circulars is $ 2 and 
for each succeeding one hundred $0.50. Express the cost y in dollars of 
X hundred circulars. Draw the graph of the function and determine 
from the graph the cost of printing 475 circulars. What does the change 
ratio of the function y express in this case ? Ans. y = ^ x + f . 



Ill, § 51] LINEAR FUNCTIONS 71 

51. The General Linear Function mx -h b. The change ratio 
of every function of the form mx + b is constant. 

Proof. Let (x^ , yi) and (ajg , 2/2) be any two pairs of cor- 
responding values. Then 

2/1 = w^i + & and 2/2 = mx2 -\- h ; 
hence , ^ 

2/2 - .Vi = m (icg - xi), 
I.e. 

Conversely, if the change ratio of the function y of x is con- 
stant and equal to m, the function has the form y = mx + b. 
Let (xi, 2/1) be a particular pair of corresponding values and 
(x, y) any other pair of corresponding values. By hypothesis 
the change ratio is equal to m ; i.e. 

X — Xi 

or 

y = mx — mxi -f 2/1 ; 

but — mxi 4- 2/1 is a constant, say b. Hence 
y = mx -h b. 

Hookers law affords an excellent illustration of the above theorem. 
This law states that the length 1/ of a piece of wire under tension is equal 
to its original length 6, plus the stretch, which is proportional to the force 
X causing it. Thus, y z=l h + mx. 

This law may also be stated simply by saying that the change ratio of 
the length y^ with respect to the pull a:, is constant. 

The preceding considerations lead to the following theorem. 

Theorem. If a function y of a variable x is such that any 
change in the value of the function is always equal to m times the 
corresponding change in the variable, the function y is given by 
a relatian of the form y — mx -f- b, and, conversely, in any func- 
tion of this form any change in y is always m times the corre- 
sponding change in x. 



72 



MATHEMATICAL ANALYSIS 



[III, § 52 



52. The Graph of a Linear Function. Let Pi{xi , y^) be any 
point on the graph of a linear function (Fig. 32). From Pj draw 
to the right a positive horizontal segment P1Q2 equal in length 
to X2 — Xi , i.e. Ax. Through Q2 draw a vertical segment and let 
it meet the graph in the point Po • The segment Q2P2 is equal 
to 2/2 — yi? i'^- ^y, and is positive if P2 is above Pi (Fig. 32 a) 
}' . __ 



Pz 



Pi 



Q% Qs 



la) 



X 

Fig. 32 



Pi 



Q2 Qb 



A 



A 



(6) 



and negative if P2 is below Pi (Fig. 32 6). Now let us take 
another positive change Ax = x^ — Xi, represented by P] Q.^ and 
the corresponding change Ay — y^ — 2/1 represented by Q3P3 . 
If the change ratio is constant, then (1) either Pg and P3 are 
both above Pi or they are both below Pj , according as the given 
constant is positive or negative; and (2) the triangles P1Q2P2 
and P1Q3P3 are similar. Therefore the points P1P2P3 are on a 
straight line, if and only if the change ratio is constant. 

Theorem. The graph of any function of the form y = mx -f- b 
is a straight line. 

To draw the graph of such a function we need, therefore, 
merely to plot two points of the graph and draw the straight 
line through them.* 

* While two points are sufficient to determine the line completely, it is 
desirable to find a third point as a cheek on the other two. Moreover, it is 
advisable to take the points as far apart as convenient. Why ? 



Ill, § 54] 



LINEAR FUNCTIONS 



73 





Y 




C 


n 






B 


y^ 




D 




,^ 


1 




y 







IM 


2 X 



Fig. 33 



In the figure this ratio 



Example. Draw the graph y = ^x -\- 2. We notice that (0, 2) and 
(4, 14) are two points on the graph. The line joining these two points is 
the required line. Check by plotting a third point. 

53. The Slope of a Straight Line. The graph of the func- 
tion y = mx + b may be obtained by observing that x = Oj 
y -^b and x = 1, y = m -\- b are 
two pairs of corresponding values 
of X and y. In the adjoining 
figure (Fig. 33) we have plotted 
the two points B (.0, b) and 
0(1, m + b) on the assumption 
that both of the quantities b and 
m are positive numbers. The 
change ratio, as we have seen, is m. 
is DC/BD. 

Now suppose that b remains constant and that m takes on 
successively different values. Under the hypothesis that b 
and BD remain fixed, the points B and D would remain fixed 
and the point C would move up or down on the vertical line 
through D, according as m increases or decreases. The line 
BG would then rotate about the point B, becoming steeper if 
m is increased and less steep if m is decreased. The change 
ratio m then measures the steepness of the line. The term 
change ratio applies to the function mx -\- b ; when applied to 
the straight line y = nix -f 6, it is called the slope of the 
straight line. 

54. Remarks Concerning the Slope of a Line. We as- 
sumed in the last section that both b and m were positive 
numbers. Let us now suppose that b is still positive, but that 
m is negative. Observe that in the preceding figure MC 
= MD + DC = b-\-m. Eecalling that the relation MC= MD 
+ DC holds universally for any three points M, D, C, on a 



74 MATHEMATICAL ANALYSIS [III, § 54 

line (Art. 35), the interpretation of a negative m, i.e. DC, is 
that the point C is below, the point D. (Cf. also § 52.) A 
negative value of m then merely causes the line to slope 
downward in going from left to right, while, as we have seen, a 
line with a positive m slopes upward. When m = 0, the line 
is parallel to the a^-axis. Indeed the equation y = mx + h be- 
comes, for the value m = 0, the equation y = h. This equation, 
when interpreted as a function of x, means that for every value 
of X, the value of ?/ is h \ the graph of such a function is ob- 
viously a straight line parallel to the avaxis. Since a change 
in X in this case produces no change in y, the change ratio is 
zero. Finally, if h is negative, nothing is changed except that 
the point B is below the origin 0. A positive m still indicates 
an upward slope and a negative m a downward slope, in pass- 
ing from left to right. 

The number h, we have seen, represents the segment from 
the origin to the point in which the line cuts the y-axis. This 
segment is called the y-intercept of the line. Similarly, the 
segment from the origin to the point in which the line cuts the 
ic-axis is called the x-intercept of the line. 

We have then the following results : 

Tlie straight line represented by the equation y = mx + b has a 
slope equal to m a7id a y-intercept equal to b. In passing from left 
to right, the straight line slopes downward ifm is negative and up- 
ward if m is positive; if m is zero, the line is parallel to the x-axis. 

In the terminology of functions we have : 

The linear function mx -f- b is an increasing function of x {i.e. 
the function increases as x increases) if the charige ratio m is pos- 
itive, and a decreasing function of x (i.e. the function decreases 
as x increases) if m is negative. It is a constant function if m 
is zero. 



Ill, § 55] 



LINEAR FUNCTIONS 



75 



55. Examples of Linear Functions. Example l. On a Fah- 
renheit thermometer the freezing point of water is placed at 32°, the boil- 
ing point at 212°. On a Centigrade thermometer the freezing point is 
at 0°, the boiling point at 100^. Express the temperature of y° Fahren- 
heit as a function of y° Centigrade. 

Solution : y = 32 when a; = 0. Also the range of temperature from 
the freezing point to the boiling point of water is 212°-32° or 180° F. while 
it is 100° C. Therefore it follows that an increase of 1° C. is equivalent 
to an increase of | of a degree F. Now as the temperature increases from 
0° to ic° C. the change in the number of degrees is x. This change in 
temperature is equivalent to an increase from 32° to y° F. The change 
in the number of degrees is then 

?/-32i=|x, or y =|x + 32. 

As a check we may observe that, when x = 100, the formula gives y = 212, 
as it should. Are negative values of x admissible ? Figure 34 represents 
the graph of this function. It was drawn by using the points A (-30, -22), 
B (100, 212). [Why is it desirable to choose points so far apart?] 
This "graph may be used to read off without computation the approximate 
temperature in F. for a given temperature in C. For example, to x = 22 
corresponds y = 72, approximately. Therefore 22° C. is equivalent to 
about 72° F. By computation we find that y = ll.Q. 





■ 










1 






' 




' 




" 




' 












. .? 
















zko 




























i 
















































^ 
















































' / 






^ 










































t 












































/ 


















'4 


1 





























/ 


















js 




























/ 




















^ 


























/ 






















*- 
















































;5 
















































1 


Si) 












































2^ 


















/ 






























y 
















/\ 














































/ 














































/ 















































/ 














































/ 














































/ 














































/ 














































/ 








































.-i 


5 



































































p'fl 








no 






7 a 


. 










/ 






















I) 


e. 


rj. 


es 


4 


iti 


I' 


ii.(.e\ 




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.- 


1 1 









Fig. 34 



76 MATHEMATICAL ANALYSIS [III, § 55 

Example 2. A bar of iron 3 ft. long at 60"^ F. will expand or con- 
tract if tlie temperature increases or decreases. The increase in length is 
proportional to the increase in temperature (physical law). More pre- 
cisely, an increase of 1° F. produces an increase of 0.0000027 ft. In this 
case we have m = 0.0000027. If y represents the length at x° F., we have 

?/ = 3 + m(x-60). 

Does this relation hold also when x < 60 ? Why ? Can you draw the 
graph ? 

We shall now give an example in which m is negative. 

Example 3, An aeroplane starts 200 miles eas^of Chicago and travels 
towards Chicago. Express its distance y from Chicago in miles at the 
end of t hours, if the aeroplane moves at the rate of 82 miles per hour. 

Solution : According to the data the distance from Chicago is de- 
creasing at the rate of 82 miles per hour, i.e. m =— 82. Therefore, 

y-2Q0=-82t, or y = -S2 t -\- 200. 

Draw the graph. What is the significance of a negative y (e.g. when 
^ = 5)? When does the aeroplane reach Chicago ? When is it at a point 
63 miles east of Chicago ? When is it 52 miles west of Chicago ? How 
could these questions be answered from the graph alone ? 

EXERCISES 

1. On a Reaumur thermometer the freezing point of water is at 0°, 
the boiling point at 80°. Express the temperature in degrees Fahrenheit 
in terms of the temperature in degrees Reaumur. Draw the graph and 
show how it may be used. 

2. Is there any temperature whose measures in the Fahrenheit and in 
the Centigrade scales are equal ? Answer by computation. How could 
the result be found graphically ? 

3. A cistern that already contains 300 gallons of water is filled at the 
rate of 50 gallons per hour. Show that the amount of water y in this cis- 
tern at the end of x hours is y = 50 x -}- 300. Draw the graph and discuss. 
How would the function be changed, if the cistern were being emptied 
at the rate of 50 gallons per hour ? 

4. A tank contains 16 gallons of water. A faucet is opened which 
admits 4 gallons per minute. Express the amount, w, of water in the 
tank at the end of t minutes. Draw the graph. Do negative values of t 
have any significance ? When will the tank contain 37 gallons ? 



Ill, §56] LINEAR FUNCTIONS 77 

5. A tank containing 37 gallons of gasolene is emptied at the rate of 
5 gallons per minute. Express the amount of gasolene in the tank at the 
end of t minutes. Draw the graph. When will the tank be emptied ? 
For what range of values of t has the function any significance ? 

6. On a certain date a man has 1 5 in the bank. At the end of every 
week he deposits $3. How much money has he in the bank at the end 
of X weeks ? Draw the graph of this function. How is the rate of in- 
crease shown in the graph ? 

7. On a certain day a man has 1 100 in the bank. At the end of 
every week he draws out '$5. How much money has he in the bank at 
the end of x weeks ? Draw the graph of this function. How is the rate 
of decrease shown in the graph ? 

8. In experiments with a pulley, the pull P in jjounds required to lift 
a load L in pounds, was found to be P = 0.15 L + 2. Plot this relation. 
How much is P when L is zero. How much is P when L is 10 lbs.? 

9. If h represents the height in meters above sea level, and h the 
reading of a barometer in millimeters, it is known that b — k ■{■ hm, where 
k and m are constants. At a height of 110 meters above sea level the 
barometer reads 750 ; at a height of 770 meters it reads 695. What 
equation gives the relation between b and Ji ? Draw the graph of this 
equation and from the graph determine h when b = 680. 

56. Linear Interpolation. The fact that the change Ay in a 
linear function y is proportional to the change Aa; in the variable 
X makes it possible to interpolate readily. For example, if we 
know that ?/ is a linear function of x, and that y = 432.50 when 
X = 32.0 and that y — 436.90 when x = 33.0, we can calculate 
mentally the value of y when x = 32.3. For we know that in 
this case Ay = 4.40 when Ax = 1.0 ; hence A?/ = 4.4 x 0.3 = 1.32 
when Ail' = 0.3. Hence y is 433.82 when x = 32.3. This pro- 
cess is known as linear interpolation. Why would this process 
not apply directly to functions that are not linear ? 

EXERCISES 

Assuming that y is a linear function in each of the following cases 
compute the values of y indicated. 

1. When a; = 10, ?/ = 50; when x = 14, y = 90; when a; = 11, y = ? 

2. When x = 2.4, ?/ = 9.8 ; when x = 2.5, y = 8.Q; when x = 2.42, y=? 



78 



MATHEMATICAL ANALYSIS 



[III, § 57 



57. Graphic Solution of Problems. Whenever we know 
at the outset that the solution of a problem is going to depend 
on the consideration of one or more linear functions, we can 
often solve the problem graphically without determining these 
linear functions analytically. Such a method is advantageous 
whenever the computation is difficult or tedious and when 
great accuracy is unnecessary. In order to decide whether 
the functions involved are linear or not, we usually have re- 
course to the theorem (§ 51) that, whenever the change in 
the function is proportional to the change in the variable, the 
function is linear. This is true, for example, in all cases of 
motion at a constant speed on either a straight or curved 
path ; the distance is then a linear function of the time. 

The following example will serve to illustrate the graphic 
method of solution. 

Example. At 7 a.m. a man starts to go up the 7-mile carriage road 
of Mt. Washington. At 9 o'clock he passes a party of ladies coming 
down. He reaches the top at 10 o'clock and, finding no view, he immedi- 
ately sets out on the return trip, which takes 1 f hrs. As he reaches the 
hotel from which he started he notices the party of ladies just arriving. 
At about what time did the ladies leave the top, assuming that the man 
kept up an approximately constant rate of speed on the way up and the 
ladles on the way down ? 

To solve the problem, we represent on a horizontal axis the time, mark- 
ing the hours 7, 8, 9, 10, 11, 12 and on the vertical axis the distances 

from the hotel at the foot of the 

t> m — I I ' I — I.I I I I I 

mountam. The graph' of the man 

going up the mountain is a straight 

line starting at O (at 7 a.m. he was 

at the hotel) and ending at a point 

A representing 10 o'clock and 7 

miles from the hotel. Regarding 

the ladies, we know that the graph 

of their descent is also a straight 

line. At 9 o'clock they were the 

same distance from the hotel as 

the man. The point B on the line OA, corresponding to 9 o'clock, 



/> 












A 












N 








/ 
















\ 


>v 


/ 




















> 


iP 


















/ 


/ 




\ 
















/ 








\ 










z 


/ 












\ 




















\ 


b£ 





07 



9 10 
Fig. 35 



11 12 houna 



Ill, § 58] 



LINEAR FUNCTIONS 



79 



is then one point of the ladies' graph. Another point is the point C 
at distance from the liotel at n:45. The line BC is then drawn 
and extended to Z>, representing 7 miles distance from the hotel. It is 
seen that the ladies started at about 7:30. How far was the man 
from the top when he met the ladies ? 

58. Sum of Two or More Functions. Let mix -h h^ , 

m^x-\-h2 , '", m^x -f- 6^ be any k linear functions of x. The sum 
of these functions is (miX + 6]) -\-{m2X -[- 62)+ ••• H-(^fc» + &*) 
and this is equal to 

(mi+W2-f ••• +mj)x+{hi 4-62H h &jk)> 

which is again of the form mx + h. The result may be stated 
as follows : The sum of any number of linear functions of x is 
itself a linear function of x. 

Example. An empty tank is being filled by a faucet supplying 2 
gallons of water per minute. After this faucet has been running 5 
minutes a second faucet is turned on which supplies water at the rate of 
3 gallons per minute. When the two faucets have been running to- 
gether for 6 minutes, an outlet is opened, but both faucets continue to 




Fig. 36 



run. If the tank is empty at the end of 32 minutes, counted from the 
start, draw a graph representing the amount of water in the tank at any 
instant. Find approximately the rate of flow from the outlet, which may 
here be considered constant. 



80 MATHEMATICAL ANALYSIS [III, § 58 

We shall represent minutes on the horizontal scale and gallons of water 
in the tank on the vertical scale. The increase of water due to each 
faucet is at a constant rate, and the decrease when the outlet is opened is 
also at a constant rate. The amount of water in the tank due to each 
cause separately is, therefore, a linear function of the time, and their 
algebraic sum is also a linear function of the time. The first faucet begins 
at < = (when w, the amount of water in tank, is 0) to supply water at 
a uniform rate which would supply 40 gallons in 20 minutes. The 
amount of water in the tank due to the first faucet almie would then be 
represented at any instant by the straight line OA joining the points 
O (0, 0) to A (20, 40). The second faucet begins when t = 5 to 
supply water at the rate of 30 gallons in 10 minutes. If this second 
faucet were operating alone, the water supplied by it at a given instant 
would be represented by the straight line joining B (5, 0) to C (15, 
30). In the actual problem from the instant i? = 5, the two faucets are 
running simultaneously. The sum of the two functions is then rep- 
resented by the line-segment DE, where Z> = (5, 10) and £' = (10, 20 + 15) 
= ( 10, 35). This line may be obtained graphically from the figure. When 
t = 11, a new factor enters, which reduces the amount of water in the 
tank to zero at i = 32. You may now finish the discussion. The required 
graph is the broken line ODHI. What would be the effect on the graph 
if one or both faucets were turned off at < = 20, the outlet remaining open ? 

EXERCISES 

1. A man on horseback rides from a place J. to a place 5, 15 miles 
distant, in 2 hours. When he is 4 miles from A, he passes a lady walk- 
ing in the same direction. The man remains at B \ hour and then 
returns to A on foot. After walking 1 hour, he meets the lady on her 
way to B. If the man walks at the rate of 3 miles per hour, find the 
rate at which the lady is walking and at what time she left A. 

2. A man starts at A to walk through B to a place C. At the same 
time a second man starts to walk from B to C. The first man reaches 
B in 1^ hours, while the second man has only walked f as far in this 
time. In how many hours will the first man overtake the second ? 

3. Represent graphically on the same drawing the motion of the hour 
and the minute hand of a clock and use the drawing to determine ap- 
proximately at what time the two hands are in the same position. 

[Hint: The hands are together at twelve. Lay off the hours from 12 
(or, 0) to 12 on the horizontal axis and the angles in degrees that either 



Ill, § 59] 



LINEAR FUNCTIONS 



81 




hand makes with the 12 o'clock position on the vertical axis. Each liand 

moves at a constant angular speed. The graph of the hour hand is then 

a straight line joining the points 

(0, 0), (12, 360). The minute hand 

goes from to 360 in 1 hour. The 

graph during the first hour is then a 

straight line joining (0, 0) to (1, 360). 

At 1 o'clock the graph begins at 

(1, 0) and goes to (2, 360) and so on.] 

4. At what time between five and 
six o'clock are the hands of a watch 
together ? 

5. At what time between two and three o'clock are the hands of a 
watch opposite to each other ? At right angles ? 

6. At what time between four and five o'clock are the hands of a 
clock at right angles ? (Two solutions.) 

7. A and B start to walk towards each other from two towns 15 
miles apart. A walks at the rate of 3 miles per hour but rests one hour 
at the end of the first 6 miles. B walks 4 miles per hour but rests two 
hours at the end of the first 4 miles. In how many hours do the two men 
meet ? 

8. Two men can do a certain piece of work in 12 and 15 days re- 
spectively. After the first man has worked 3 days alone, the two men 
finish the work. How long do they work together ? Ans. 5 days. 

9. A messenger boy riding a bicycle at the rate of 9 miles per hour 
is sent to overtake a man on horseback riding 6 miles per hour. How 
long will it take the boy to overtake the man if the man had a start of 
4 miles ? 



59. Explicit and Implicit Functions. We have hitherto 
considered functions which were defined explicitly by an 
expression involving the variable. Thus the relation between 
y° Fahrenheit and x° Centigrade was expressed by the relation 



Now let us consider the equation 2x — Sy-{-7 = 0. This 
equation also defines a functional relation between two vari- 



82 MATHEMATICAL ANALYSIS [III, § 60 

ables. To every value of x corresponds a definite value of 2/, 
and, conversely, to every value of y corresponds a definite 
value of X. But, the equation does not express one of the 
variables explicitly as a function of the other. In fact the 
form of the equation gives no indication which of the variables 
is to be considered as the independent variable and which as 
the function. Such a relation is said to define a function 
implicitly. 

From such an implicit relation we can derive the expression 
of either variable as an explicit function of the other. Thus 
from 2iB— 32/-f7 = follows at once 

2/ = l^+i and x=:f2/-f 

The first of these equations expresses y as an explicit function 
of X, and the second expresses x as an explicit function of y. 

60. The General Equation i4x -h 5y + C = 0. Any linear 
relation between two variables x and y can be written in the 
form 
(1) Ax^-By-\-C^Qf. 

For example, the relation just discussed in the preceding arti- 
cle is obtained from this general relation by placing ^ = 2, 
5 = — 3, (7=7. Equation (1) always defines ?/ as a linear 
function of x^ except when 5 = 0. In this case the term in- 
volving y drops out and the equation reduces to Ax + C = 0, 
and we cannot speak of 2/ as a function of x. 
But, if 5 :^ 0, we have By = —Ax—C^ or 

A C 

y = — ^ — J 

^ B B' 

which is of the form y = mx -f h. Hence we conclude : 

Any equation of the form Ax + By -\- C =0 defines y as a 
lineal' function of x for all values of A, B, C except B = 0. 



Ill, § 61] LINEAR FUNCTIONS 83 

61. The General Equation of the Straight Line. It follows 
from the result of the last section, that the locus of the equa- 
tion 

when interpreted geometrically in rectangular coordinates, is 
a straight line, except perhaps when B = 0, when the equa- 
tion takes the form Ax -\- C = 0. In this case, if A = also, the 
equation reduces to (7 = 0, and it completely disappears. If A 
is not zero, we may solve the equation for x and obtain, 

G 
. = --, 

or 

« = a constant. 

Now, the locus of a point whose abscissa is constant is a line 
parallel to the ?/-axis and at a distance equal to the constant 
from it. Thus the locus of a; = — 3 is a line parallel to the 
y-Sixis, and three units to the left of it. 

The case JB = is not then an exception, and we have the 
following theorem. 

Every equation of the form Ax ■\- By -\- C = 0, when repre- 
sented geometrically by means of rectangular coordinates, repre- 
sents a straight line. If B = 0, the line is parallel to the y-axis ; 
1/^ = 0, the line is parallel to the x-axis; if (7=0, the line passes 
through the origin. 

Prove the last two statements of this theorem. 

We may also state the following theorems. 

Every straight line in the plane may he represented by an equa- 
tion of the form Ax -j- By -\- C = 0. 

The loci of Axi-By-{- C=0 and k (Ax -\- By -\- C) = 
(k =^ 0) a?'e identical. 

The proofs of these theorems are left as exercises. 



84 MATHEMATICAL ANALYSIS [III, § 62 

62. Analytic Geometry. We have thus far used the notion 
of coordinates to give a geometric interpretation to algebraic 
relations. It is possible to reverse the process and use the 
connection established between algebra and geometr}^, for the 
study of geometry. This method of studying geometry by 
algebraic means is called analytic geometry. In the following 
sections we proceed to develop certain analytic methods 
applicable to the straight line. The results are, in a large 
measure, merely a restatement from a different point of view 
of the results already obtained. 

63. Straight Lines. We have already seen that the graphs 
of equations Ax -\- By -\- C = and y = mx + & (§ 52), when 
represented by means of rectangular coordinates, are straight 
lines. In § 60 we saw that the first of these equations could 
be put in the form of the second, provided 'B ^ 0. Thus when 
an equation of the form Ax -{- By -\- C = is solved for y, the 
coefficient of x is the slope, and the constant term is the y-intercept. 

The slope of the line connecting the two points Pi(iCi, 2/i), 
A(aJ2,2/2) is (§§51-53) 

We see geometrically that a line is determined when we 
know its slope and a point on the line. To determine the 
equation of this line, if {x^, y^) is the given point and m the 
given slope, we proceed as follows. Let {x, y) be any variable 
point on the line. Then, equating slopes, we have 

X — Xi 

that is, 

y-y,= m{x- X,) 

is the required equation. 



Ill, § 64] 



LINEAR FUNCTIONS 



85 



It is left as an exercise to prove that the equation of the 
straight line through the two given points (x^, y^), (ajg, 2/2) is 



if Xi =^ X2. 



2/-yi = ^'_^' (^-^i)> 



64. Parallel Lines. In Fig. 37 let (1) and (2) be two 
parallel lines with slopes mi and mg. Construct the positive 
segments PiQi and P2Q2 from the points P^ and P2 on lines (1) 




Fig. 37 



and (2) respectively, and complete the right triangles P^QiRi 
and P2Q2R2' We then have 



m 



, = ^1^ and m2 = ^'^' 



If the lines are parallel, QiRi and Q2R2 are either both positive 
or both negative ; m^ and mg have then the same sign. They 
have the same magnitude since the triangles P^QiRi and 
P2Q2R2 are similar. Hence, 

If two lines are parallel, their slopes are equal, i.e. m^ = mj. 
Conversely, if the slopes of two lines are equal, the lines are 
parallel. 

The proof of this statement is left as an exercise. 



86 



MATHEMATICAL ANALYSIS 



[III, § 65 





7 


V 




{ 

9 






A 




y 







\ 


\ 


X 



Fig. 38 
vertical line B1R2 . 



65. Perpendicular Lines. In 

Fig. 38 let (1) and (2) be two 
perpendicular lines with slopes 
mi and rriz and let the units on the 
two axes be equal. From the 
intersection P of the two lines 
construct the positive horizontal 
segment PQ of any convenient 
length. Through Q draw the 



We then have 



m, = 



^Ma„dm, = «^ 



PQ 



PQ 



Therefore the slopes have opposite signs. Why ? Also from 
the right triangle R.R^P we have P^ = | QE^ \ • | QR2 1. There- 
fore * mim2 = — 1 and 

m, =- 






That is, if the units on the coordinate axes are equal , perpendicu- 
lar lines have slopes ivhich are negative reciprocals of each other. 
Conversely, if the slopes of two lines are negative reciprocals of 
each othery the lines are perpendicular, provided the units on the 
coordinate axes are equal. The proof of this statement is left 
as an exercise. Why is it necessary to assume the units equal ? 



66. Illustrative Examples. Example 1. Find the equa- 
tion of the straight line through the point (4, 7) and having 
the slope — 2. 

* This proof presupposes that neither mj nor mg is zero, i.e. the lines are 
not parallel to the coordinate axes, and the result obtained does not apply to 
such lines. However, two lines parallel to the x- and y- axes have equa- 
tions of the form y = a constant and a; = a constant, respectively, and henco 
can be recognized at once. 



Ill, § 66] LINEAR FUNCTIONS 87 

We have at once from § 63, ?/ — 7= — 2 (ic — 4) 
or 

2 a; -f 2/ - 15 = 0. 

Example 2. Find the equation of the straight line through 
the points P^{2, 4), P^{- 5, 6). 

rru 1 6-4 2 

The slope m = - — - — - = 

— 5—2 7 

From § 63 the equation of the line is y ~ 4:= — ^ (x — 2) ov 
7y + 2a;-32 = 0. 

Example 3. Express the temperature measured by y° Fah- 
renheit as a function of fl;° Centigrade. 

We know that when y = 32, x = 0: i.e. Pi(0, 32) is a point 
on the graph. In the same way we have PaC^^O, 212) a point 
of the graph. Therefore the equation of the line connecting 
these points is 

' 212-32 ^ y-32 
100 - a; - 
or 

2/ = 9 a; + 32 (See § 55, Example 1.) 

Example 4. Find the equation of the straight line 
through the point (2,-5) and parallel to the line 2 y + 4 ic — 5 
= 0. 

The slope of the given line is — 2 (§ 63). Therefore 
the equation of the required line is y-{-5= — 2(x — 2) or 
2a; + 2/ + 1 = 0. 

Example 5. Find the equation of the straight line through 
the point (1, — 2) and perpendicular to the line 3x — y -\-2 = 0. 

The slope of the given line is 3. Therefore the slope of 
the required line is — ^ (§ 65). The equation of the required 
line is2/-f2 = — i(a; — l)ora;-f32/4-5 = 0. 



88 MATHEMATICAL ANALYSIS [III, § 66 

EXERCISES 

1. What is the meaning of the constants m and h in the equation 
y = mx + b? 

2. What is the effect on the line y = mx + & if 6 is changed while m 
remains fixed ? If m changes when b remains fixed ? 

3. Describe the effect on the line y — yi = m(x — xi) if m changes 
while xu y\ remain fixed : also describe the effect if Xi, yi, vary while m 
remains fixed. 

4. What is the equation of the line 

(a) whose slope is 3 and whose ^-intercept is 2 ; Ans. y = 3 x + 2. 
(6) whose slope is 4 and whose y-intercept is — 3 ; 

(c) whose slope is and whose ^/-intercept is — 1 ; 

(d) whose slope is and whose y-intercept is ? 

5. Describe the positions of lines (c) and (d) in Ex. 4. 

6. Define " y-intercept of a line." What is meant by the "x-inter- 
cept"? 

7. For each of the following lines give x-intercept, y-intercept, and 
slope : 

(a) 2x-3?/ = 7. ^ns. I; -I; f. (c)2x-y + 5 = 0. 

(6) X + y - 2 =: 0. ((?) 4 X + 2/ = 0. 

8. Is a straight line determined if we know its intercepts ? Try 
each of the equations 2 x — y = 4 and 2 x — y = 0. 

9. Find the equation of the line joining the two points (2, 1) and 
(—3, 1); of the line joining the points (4, 2) and (4, —3). 

10. Which of the following lines are parallel ? 

(a) 2 X - 2/ - 4 = 0. (c) 4 X - 2 ?/ - 1 = 0. 

(6) y + 2 X + 3 = 0. (d) 2 y + 4 X + 5 = 0. 

11. Are the points (1, 5), (- 1, 1), (2, 6) on the line y = 2 x + 3 ? 

12. What is the equation of the line which is parallel to y = 2 x + 3 
and passes through the origin ? perpendicular to y = 2 x + 3 and passes 
through the origin ? 

13. Determine k so that 

(a) the line 2x + 32/ + A; = shall pass through the point (0, 1) ; 

Ans. - 3. 

(b) the line 2x-f-3y4-* = shall have a y-intercept equal to 2 ; 
{c) the line 2x + 3y + ^ = shall have an x-intercept equal to 5. 



Ill, § 66] LINEAR FUNCTIONS 89 

14. Find the equations of the sides of the triangle whose vertices are 
(3,4), (-1, 2), (-4, -5). 

Ans. x-2y + 5 = 0; 9x-ly + l=0- 7x-3i/+13 = 0. 

15. Find the equations of the sides of the quadrilateral whose vertices 

are (-2,1), (3, -1), (- 2, 4), (1, 7). 

16. What intercepts does the line through the points (2, — 7) and 
(4, — 5) make on the axes ? 

17. Find the equation of the line which passes through the point 
(4, — 2) and whose slope is 6. 

18. A line has the slope 2 and passes through the point (—1, 2). 
What are its intercepts ? 

19. What is the equation of the line which passes through (—6, 5) if 
its y-intercept is — 3 ? Ans. 8a: + 5?/ + 16 = 0. 

20. Write the equations of the lines which make the following inter- 
cepts on the X- and !/-axes. 

(a) 2 and - 4 ; (b) - 7 and - 3 ; (c) 4 and 5 ; (d) and 0. 

21. If the X- and y-intercepts of a line are a and b, prove that the equa- 
tion of the line can be written iii the form 

a b 

[This equation is called the intercept form of the equation of a straight 
line.] 

22. Solve Ex. 20 by using the result of Ex. 21. Does the formula hold 
in Ex. 20, (d) ? Explain. 

23. Find the equation of the straight line through the point (4, — 6) 
parallel to the line 2x— y-{-7 = 0; through the same point, perpendicu- 
lar to the line 2x — y + i = 0. Ans. y = 2x— IS; 2y = — x — Q. 

24. Prove that the lines Ax + By -\- C =0 and Ax -{- By + D = are 
parallel. State this theorem in words. 

25. Prove that the lines Ax -\- By + (7 = and Bx—Ay-^D = are 
perpendicular. State this theorem in words. 

26. Prove that the lines Ax -\- By + = and Mx -h Ny -\- P = Sive 
perpendicular if and only if AM -{■ BN = 0. 

27. Show that the points (- 8, 0), (-4, - 4), (-4, 4), and (4, - 4) 
are the vertices of a trapezoid. 

28. The Reaumur thermometer is graduated so that water freezes at 0° 
and boils at 80°. Find the equation of the line that represents the read- 



90 



MATHEMATICAL ANALYSIS 



[III, § 66 



ing B of the Reaumur thermometer as a function of the corresponding 
reading C of the Centigrade thennometer. 

29. A printer asks 75 cents to set the type for a notice and 3 cents per 
copy for printing. The total cost is what function of the number of 
copies printed ? Draw the graph of this function. 

30. Express the value of a $ 1000 note at 6 % simple interest as a 
function of the time in years. Is this a linear function ? 

31. A cistern is supplied by a pipe that supplies water at the rate of 30 
gallons per hour. Assuming that the amount A of water in the cistern 
is connected with the time ^ by a linear relation, find this relation if 
^ = 1000 when t = 10. What is A when t = 0? 

32. In stretching a wire it is assumed that the elongation e is con- 
nected with the tension t by means of a linear relation* Find this rela- 
tion if i = 20 lb. when e = 0.1 in. and t = 60 lb. when e = 0.3 in. 

67. Systems of Straight Lines. An equation of the first 
degree in x and y, and containing an arbitrary constant, repre- 
sents in general an infinite number of 
straight lines. For the equation will 
represent a straight line for each value 
of the constant. All the lines repre- 
sented by an equation of the first 
degree containing an arbitrary con- 
stant are said to form a system of 
lines. The arbitrary constant is called 
the parameter of 
the system. 
Thus the equa- 
tion yz= — 3x-{-b represents the system 
of straight lines with slope — 3. (See 
Fig. 39.) The equation y — 2 = 7n{x — l) 
represents the system of straight lines 
through the point (1, 2).* (See Fig. 40.) 

* It represents every line of this system except the one parallel to the 
y-axis. Why ? 




Fig. 39 





1 




c 


: i^.Z-t 




^v 


i X 


- - ^:,: 






[V i y 


' 


sA' *" 




J^S 


r-t 






-Q"~ ^ V 


:.^^^\n. 


i :: -5^ 




ii 1i-^ X 


::: dZ 


±i: :± 



Fig. 40 



lU, § 68] LINEAR FUNCTIONS 91 

68. Pencil of Lines. All the lines in a plane which pass 
through a given point are said to form a pencil of lines. The 
point is called the center of the pencil. 11 Ax + By ~\- C = 0, 
and A'x + B'y -\- C' = are any two lines of the pencil, then 

(3) (Ax -f % 4- 0)-h k(A'x + B'y + 0')= 0, 

where k is an arbitrary constant, represents a line of the 
pencil. This is true because the equation- (3) 

(a) is of the first degree in x and y and therefore represents 
a straight line ; 

(b) is satisfied by the coordinates of the point of intersec- 
tion of the two given lines. Why ? 

Example 1. Eind the equation of the line through the 
point (2, — 5) and parallel to4:X-{-2y-{-5 = 0. 

The system of lines parallel to 4:X -\-2y -\- 5 = is given 
by the equation y= — 2 x -\-7c. Since we want the particular 
line of the system that passes through the point (2, — 5), the 
equation must be satisfied by these coordinates. It follows 
that, — 5 = — 4-f-A:orA;= — 1. 

Therefore, y = —2x — 1 is the desired equation. 

Example 2. Find the equation of the line through the 
point (4, — 1) and perpendicular toSx-\-2y — 5 = 0. 

The system of lines perpendicular to 3x-\-2y — 5 = is 
given by the equation y = ^ x -\- k. Since we want the line of 
the system that passes through the point (4, — 1), we have 
k = — y . Therefore, the desired equation is 

y=2x_i_i or 2a; -31/ -11 = 0. 

Example 3. Find the equation of the line through the 
intersection of 2x-\-y — 4: = and x-{-y—l=0, and perpen- 
dicular to x-\-2y = S. 

Any other line through the intersection of the given lines is 

(4) (2x -{- y - 4:)+ k (x + y - 1) =0 



92 MATHEMATICAL ANALYSIS [III, § 68 

or 

x{2 + k)-\- y{l + k) + {- i - k) = 0. 

The slope of this line is — (2 + k)/{l + A:) and this must 
be equal to the negative reciprocal of the slope of the straight 
line X -f 2 2/ = 3. Therefore, 

-2^^=2 and k=-^. 
1 + k 3 

Substituting this value in equation (4) and simplifying, we 
have 2x — y — S = 0, the required equation. 

EXERCISES 

1. Find the equation of the straight line through the point (1, 5) and 
parallel to2x-\-Sy— 9=0; perpendicular to2x + 32/ — 9 = 0. 

Alls. 2x +3?/-17 = 0; 3x-2y+ 7 =0. 

2. Find the equations of the altitudes of the triangle whose vertices 
are (2, 8), (4, - 5), (3, - 2). 

3. Find the equation of the straight line through the intersection of 
10x4-5^ + 11 = and x +2y + 14 = which is perpendicular to 
x + 7y + l=0; parallel io S x - 7 y = I. 

4. Find the equation of the straight line through the intersection of 
X -\-2y — 4: = and x — Sy-^1=0 which is perpendicular to Sx — 2y 
+ 4 = 0; parallel to x — y = 0. 

5. Find the equation of the straight line through the intersection of 
X -\-y - 1 = 0, x-3y+4 = and 

(a) through the point (1, 1) ; Ans. x + 6y — 6 = 0. 

(&) parallel to the line x + 2?/ — 9 = 0; 

(c) perpendicular to the line 4 x — 5 ?/ = ; 

((7) through the intersection of3x + 4?/ — 8 = and x— 5j/ + 7=0. 

6. Find the equation of the straight line which passes through the 
point 

(a) (0, 0) and is parallel to2x — y + 4 = 0; 

(6) (1, 2) and is perpendicular to3x— 2^/ — 1 =0; 

(c) (— 1, 2) and is parallel to x — y — i=0. 

7. Find the equation of the line which passes through the inter- 
section of X — 2^ + 2 = and x + y = and through the intersection of 
x + y+2 = 0, x-?/ = 0. 



Ill, § 69] 



LINEAR FUNCTIONS 



93 



8. Find the equation of the straight line through the intersection of 
x-2y -{-7 = and 2x-y-i-S = and 

(a) parallel to the aj-axis ; 
(6) parallel to the «/-axis. 

9. Find the equation of the straight line which passes through the 
intersection ofSx— ^ + 2 = and x + y = 6 and which 

(rt) passes through the origin ; 

(b) is parallel tox — 4?/ + 3=0; 

(c) is perpendicular to 3aj — 2^ + 4 = 0. 

69. Intersection of Two Lines. Simultaneous Equations. 

We have just seen that linear equations in one or two vari- 
ables are represented in rectangular coordinates by straight 
lines. We now wish to determine the coordinates of the point 
of intersection of two lines whose equations are given. That 
is, algebraically, we wish to find a set of values for x and y 
which satisfy both equations. 

Example 1. Solve the equations 



(6) 
(6) 



Sx — 4:y = 7. 
x-\-2y=9. 



m 





Multiplying equation (6) by 2 and add- 
ing the result to equation (5), we obtain 
5 a; = 25, or X = 5. 

Likewise multiplying equation (6 ) by 3 pj^ ^i 

and subtracting the result from equation 

(5), we have — 10 y = — 20, or y = 2. The set of values x = 5, y = 2 is 
seen to satisfy both equations and is called the solution of the given 
equations. If we plot lines (5) and (6) (Fig. 41), we see from their 
graph that the coordinates of their point of intersection are (5, 2). 

Therefore, a method of solving two linear equations in one or two 
variables is to plot the lines represented by each equation, and then deter- 
mine from the graph the coordinates of the point of intersection. The 
algebraic method of first eliminating one variable and then the other has 
the advantage over the geometric method in that it is always accurate. 
Instead of eliminating twice, the value found for either variable can be 
substituted in either equation, and the value of the second variable de- 
termined. 



94 



MATHEMATICAL ANALYSIS 



[III, § 69 



Example 2. 



Solve the equations 

(7) 
(8) 













T 




~ 






~ 




~ 






n 




- 


























C\ 


































^^ 


^ 






























^A 




^ 































•i 












% 

^ 




^' 




^ 


- 




























?- 














































gf* 


^ 














^ 


y 
















^ 












n 






















Y" 




































■" 


_ 


_ 








_ 



























x-2y = S. 

X — 2 ?/ = — 5. 



Subtracting the second equation from 
the first, we obtain = 8. Tliat is, there 
are no values of x and y satisfying both 
equations. Such equations are said to 
be inconsistent or incompatible. We see 
that lines (7) and (8) have the same slope, 
but different y-intercepts, and therefore are parallel lines. 

Example 3. Solve the equations 



Fig. 42 



(9) 
(10) 



x-y = 2. 



2x 



2y=4. 

Multiplying the first equation by 2 and subtracting the second from it, 
we have 0=0. If equation (10) be divided by 2, equations (9) and (10) 
are seen to represent the same relation between x and y, and are not 
therefore sufficient to determine x and y. We can assign to either vari- 
able an arbitrary value and then find the corresponding value for the 
other variable. The equations can, therefore, be said to have an infinite 
number of solutions. Such equations are called dependent. The graphs 
of these equations are coincident lines. 

Let us now consider the general equations 

(11) aiic + bill = Ci , 

(12) a-^x-^ b^ = C2, 

where none of the constants are zero. Eliminating y, we 
obtain {a^bz — a2&i) x = C162 — Cobi . Eliminating x, we ob- 
tain {oibo — aib^) y = aiC2 — a^c^ . Now if 0162 — 02^1 =^ ^) we 
have 

C261 



x^"^^^^ 



aiC2 — «2Ci 



we cannot 



aib2 — 0.2^1 ^1^2 ~ ^2^1 

If, however, aib^ — aa^i = 0, i.e. a.^ja^ = &2^i) 
solve for x and y. Denoting the common value of these quo- 
tients by A:, we have ag = fcai , 62 = ^^1 • Then equations (11) 
and (12) become a^x + b^y = Ci , and "ka^x -f- Icb^y = Co . 



Ill, §69] LINEAR FUNCTIONS 95 

We must now distinguish two cases according as C2 = kci or 
C2 ^fc kci . In the former case, by dividing out k, we see that the 
equations are dependent and have an infinite number of solu- 
tions. In the latter case they are inconsistent, and thus are 
not satisfied simultaneously by any values of x and y. 

Discuss the cases that arise if some of the constants are zero. 

EXERCISES 

Find, when they exist, the coordinates of the points of intersection of 
the following lines. Check your answer from a graph. 

1. 4a; + 2?/=:9. 3. x + 2y = 3. 5. a: + 4?/ = l. 
2x-5y = 0. 2x+4y = 6. 2x + 8?/ = 2. 

2. 3x + 4?/ = 12. 4. x-2y = 7. Z.x-^y = 1. 

X — y = b. 2x — iy = b, — x -\-2y = 3. 

In the following exercises are the lines concurrent ? If so, what point 
have they in common ? 

7.x + 2?/ = 3. 8.x-y = -l. 9. x+2y = 5. 10.x-2y = 3. 
x — y = 0. 2x-{-y = S. 6x— y = 3. 5 x — y = 2. 

5x— ?/=4. Sx-2y = l. 2x + y=4. 2x + 3?/ = l. 

In the following exercises, find k so that the lines shall be concurrent. 
ll.x+y = 2. 12.2x-y-0. 13. 3 x - y = 4. 

2x—y = l. x + 3y = 7. x + y = 0. 

4x + y = k. Ans. b. x -^ ky = b. b x — 2 y = k. 

14. The sides of a triangle have for their equations 2 x + y = 5, 
a;— ?/ = 10, — 2x + ?/ = 6. What are the coordinates of the vertices of 
this triangle ? What are the equations of the altitudes ? 

15. Find the equation of the straight line through (2, 1), (—1, 2), 
using the equation Ax + By + C = 0. [Hint : Solve for A/C and B/C] 

16. Find the equation of the straight line through (4, 7) and having 
the slope 3, using the equation Ax + By -\- C = 0. 

17. It has been shown experimentally, that the length I of a wire in feet 
under a tension of p pounds, is I = a + bp, where a and b are constants. 
Find a and bii 1= 190 when p = 270, and that I =190.2, when p = 450. 

18. The readings T and 8 of two gas meters are connected by the 
equation T = a -}- bS. Determine a and b when we know that 8 = 10, 
when T = 300, and S = 100, when T= 420. 



96 MATHEMATICAL ANALYSIS [III, § 69 

19. The pull in pounds to lift a load I in pounds with a pulley is given 
by the relation p =: al + b, where a and h are constants. Find a and b 
when it is known that a pull of 8 pounds lifts a load of 40 pounds, while it 
takes a pull of 2 pounds to hold the rope on when no weight is attached. 

70. Equations Containing More than Two Unknowns. It 

is easy to see that the methods employed in § 69 for solving 
a system of two simultaneous equations, each containing two 
unknown quantities, may also be employed for solving a 
system of three or more equations, involving as many unknown 
quantities as there are independent equations. 
Example. Solve the equations 

(13) 7x-\-Sy-2z = 16. 

(14) 5x-y-\- 6z = Sl. 

(15) 2x + 6y + 3z = S9. 
Adding three times (14) to (13) gives 

(16) 22 X + 13 ;s = 109. 
Adding five times (14) to (16) gives 

(17) 27 a; + 28^= 194. 

Solving equations (16) and (17) by the methods of § 69, we have a; = 2, 
z = 5. Substituting these values in (13), we obtain y = i. It is readily 
seen that x =2^ y = 4, z = b satisfies equations (13), (14), (15). 

The cases in which three simultaneous equations in three 
unknowns have no solution, or an infinite number of solutions, 
will be discussed in Chapter XXI. 

EXERCISES 
Solve the following simultaneous equations : 

l.->2x-\-4y-{-z = 12. 2. X + y + z = IS. 3. 2x -Sy - z = 2. 
^^x-\-y-z = S. x-2y-{.iz = l0. 6x-i-2y + z = -8 
x -\- y + z = 7. 3x + y~Sz=5. x—2y — z = 2. 

4. a; + 8 2/ - 4 = 9. 5. w + a- 4- ?/ = 15. 6. a: + y = 4. 
3x+3?/ — = 6. x + y -{■ z = 19,. 2x-\- z = 4. 

bx + 2y-2z = l. wj + ?/ + 0=17. y-2 = 3. 

10 + X + = 16. 



Ill, §70] 



LINEAR FUNCTIONS 



97 



7. If A and B can do a piece of work in 10 days, and A and C in 8 
days, and B and C in 12 days, how long will it take each to do the work 
alone ? 

8. Three towns A, B, and C are situated at the vertices of a triangle. 
The distance from A to B via C is 76 miles ; from A to C via B 79 miles; 
from B to C via A 81 miles. Find the distance from A to B, from B to 
C, from C to A. 

9. In a triangular track meet the following was the final score : 



SCOKE 


First Place 


Second 
Place 


Third Place 


Total 


College A . . . 
College B . . . 
College C . . . 


5 
2 

2 


3 
4 

2 


3 
1 

6 


37 
23 

22 



How many points did each place count ? 

10. Two passengers traveling from town A to town B have 500 
pounds of baggage. The first pays $ 1.75 for excess above weight allowed, 
the second $1.25. If the baggage belonged to the last passenger, he 
would have to pay $ 4 excess. How much baggage is allowed to a 
single passenger ? 

11. A crew can row 4 miles downstream and back again in 1| hours, 
or 6 miles downstream and half way back in the same time. What is 
the rate of rowing in still water, and what is the rate of the current ? 

Ans. 6 miles per hour ; 2 miles per hour. 

12. Two trains are scheduled to leave two towns A and B, m miles 
apart, at the same time, and to meet in h hours. The train leaving A 
was k hours late in starting, so the trains met n hours later than the 
scheduled time. What is the rate at which each train runs ? 

13. Two men are running at uniform rates on a circular track 150 feet 
in circumference. When they run in opposite directions, they meet every 
5 seconds. When they run in the same direction, they are abreast every 
25 seconds. What are their rates ? 

14. Find a, &, c, so that y = a -\- bx -\- cx^ shall be satisfied by (2, 1), 
(1,0), (3,-6). 

6a;2_ic-3 a 



16. Find a, 6, c, so that 



x^ 



X + 1 X, 



CHAPTER IV 

THE QUADRATIC FUNCTION 

I. GRAPHS OF QUADRATIC FUNCTIONS 

71. The General Quadratic Polynomial ax^ -h bx -{-c. 

Having considered in some detail the linear function mx + b 
and its geomietric interpretation, we now turn our attention to 
a similar study of the quadratic function, i.e. sl function ex- 
pressed by a polynomial of the second degree in one variable. 
Such polynomials are, for example, a;^ -f- 1, 100 -f- 50 ^ — 16.1 f^j 
etc. The general form of such a polynomial is ax^ -\- hx^ -f c, 
where a,h, c are constants and a^O. Such functions abound 
in practice. Thus, if a projectile be shot vertically upward 
from the top of a tower 100 ft. high, with an initial velocity of 
50 ft. per second, the distance s (in feet) from the ground at 
the end of t seconds, is given approximately by the poly- 
nomial 

s = 100-f 50i-16.1f2. 

The general formula for the distance s from the ground at the 
end of t seconds of a projectile shot vertically upward is 
(approximately) 

where Sq is the distance from the ground when t = 0, Vq is the 
initial velocity, and g is the so-called " gravitational constant," 
which varies slightly from place to place but is approximately 
equal to 32.2 when the distance s is measured in feet and the 
time is measured in seconds. 



IV, § 72] 



QUADRATIC FUNCTIONS 



99 



72. The Function x^. We consider first the simplest of all 
quadratic functions, viz. the function y = x^. A brief tabular 
representation of this function is as follows : 



X 





1 


2 


3 


4 


-1 


-2 


-3 


-4 


y 





1 


4 


9 


16 


1 


4 


9 


16 



If we plot these points, we obtain Fig. 43, in which we notice 
that the points seem to be arranged according to some regular 
law. We may insert additional points by calculating values 



~ 




Y 




















































































* 
























1 






























































































































































































































~\ 






















^///A 




















t 


'/, V' 






















z w, 






















wM 


































X 








V 





























































Fig. 43 

of y for values of x between those already used. Thus 
for X = 1.5, y = 2.25 and a; = — 1.5 y = 2.25. These points 
are also marked on the figure. In general we see that for 
X = a and also for a; = — a, we have y = a^. . Geometrically 
this means that the graph is symmetrical with respect to the 
2/-axis, i.e. if the part of the graph on the right of the y-axis 
is turned about the ?/-axis until it falls in the original plane, 
it will coincide with the part on the left of the y-Sixis. More- 
over, since x"^ is positive (or zero) for all real values of x, no 
part of the graph will be below the a;-axis. 



100 



MATHEMATICAL ANALYSIS 



[IV, § 72 



Keeping these facts in mind we shall make a more detailed 
study of this function and its graph, by considering values 
of X which are closer together. We shall confine ourselves to 
values of x between x = and x = 2. The corresponding 
values of y, for all values in this range at intervals of 0.1 of 
a unit, are given in the following table : 



X 


y 


X 


y 


X 


y 


X 


y 


0.1 


0.01 


0.6 


0.36 


1.1 


1.21 


1.6 


2.56 


0.2 


0.04 


0.7 


0.49 


1.2 


1.44 


1.7 


2.89 


0.3 


0.09 


0.8 


0.64 


1.3 


1.69 


1.8 


3.24 


0.4 


0.10 


0.9 


0.81 


1.4 


1.96 


1.9 


3.61 


0.5 


0.25 


1.0 


1.00 


1.5 


2.25 


2.0 


4.00 



We cannot, with any accuracy, insert in Fig. 43 the corre- 
sponding points of the graph. We therefore adopt a pro- 
cedure analogous to the use of a magnifying glass, in order to 
separate the points. This we have done in Fig. 44 by choos- 
ing the unit on each axis 10 times as large as in Fig. 43. We 
then see that there is no difficulty in plotting all the points 
given in the above table. 

Let us study more carefully the immediate neighborhood of 
some point on the graph, for example, P(l, 1). We shall 
magnify the shaded area in Fig. 44 in the ratio 10 : 1 and 
make use of the following table : 



X 


y 


X 


y 


X 


y 


X 


y 


X 


y 


0.90 


.8100 


0.95 


.9025 


1.00 


1.0000 


1.05 


1.1025 


1.10 


1.2100 


0.91 


.8281 


0.96 


.9216 


1.01 


1.0201 


1.06 


1.1236 






0.92 


.8464 


0.97 


.9409 


1.02 


1.0404 


1.07 


1.1449 






0.93 


.8649 


0.98 


.9604 


1.03 


1.0609 


1.08 


1.1664 






0.94 


.8836 


0.99 


.9801 


1.04 


1.0816 


1.09 

1 


1.1881 







IV, § 72] 



QUADRATIC FUNCTION-S 



101 



It will be noted that the points of the graph now lie almost 
on a straight line (Fig. 45). We have drawn a straight line 
through P for the purpose of comparison. If we should desire 
a more detailed representation in the neighborhood of the 
point P, we should calculate the values of y for values of x 
between x — .99 and x = 1.01 and draw anew a small portion 



w- 



1.2 



1.1 



1.0 



0.9 



i 



1 2 

Fig. 44 



.90 



1.0 
Fig. 45 



1.1 



of the figure about P under a tenfold increase of the unit. 
We would then find that the points would hardly be distin- 
guishable from the points on a straight line. 

Similar conclusions might be reached near any other point 
on the graph. It is of course impossible to prove this for 
each separate point by separate calculations. To prove the 
fact generally we proceed as follows. Let x^ be any particular 
value of the variable x and y^ the corresponding value of the 
function y ; then y^ = x^^. Now suppose that the value x 
is increased or decreased by a certain amount, which we shall 
call Ax (a decrease means that Ax is negative). The new 
value of X is then Xi -f Ax and the corresponding value of the 



102 



MATHEMATICAL ANALYSIS 



[IV, § 72 



function is {xi + Aa;)2, This new value of the function differs 
from the original value of the function, yi, by a certain amount 
which we shall call Ay. We then have 

2/1 + A?/ = (xi -h Axy 

= Xi_^4-2xiAx + Ax"^', 



but 



2/1 = a^i 



Therefore, by subtraction, 
or 

(1) 



Ay = 2 cciAic + Ax'^ 
Ay = (2xi-\- Ax) Ax. 



Since formula (1) is true for every value of Xi , it follows that 
Ay approaches zero when Ax approaches zero. This means 
that in the neighborhood of the point {x^ , y^) we can find new 




Fig. 46 

points on the graph whose x and y differ from those of the 
given point by as little as we please. This simply means that 
the set of all points of the graph oi y = x^ form a set of points 
with no gaps between them ; they form what we may call a 
continuous line or curve.* 

* A function is said to be continuous for a value x = Xi, if when Aa; ap- 
proaches the corresponding Ay also approaches 0. See footnote on p. 19. 



IV, § 73] QUADRATIC FUNCTIONS 103 

Further, equation (1) gives the relation, 

^ = 2 iCi + Ax (if ^x 4- 0). 
Aa; 

From the graph (Fig. 46) we clearly see that this change ratio 
is the slope of the line joining the points P^ix^ , y^) and 
P2(i>^i + Aa;, 2/1+ A?/).* If the latter point approaches the former 
along the curve, i.e. if we let Ace become numerically smaller 
and smaller, then the change ratio A?/ / Ao? will differ less and 
less from 2xy. Indeed, we may choose Aic sufficiently small 
(without making it zero) so that A?/ / Ax- will differ from 2 x^ 
by less than any previously assigned amount. 

Geometrically this means that in the immediate neighbor- 
hood of the point Pi on the graph oi y — a?^, the points of the 
graph lie very near to the straight line through Pi whose slope 
is 2 x^. From a somewhat different point of view, we can let 
the secant joining the points Pi {x-^ , y^) and Pg {xi-\-Ax, yi-\-Ay) 
on the graph rotate about Pi in such a way that Aic, and there- 
fore Ay, become smaller and smaller and the secant approaches 
a definite position. through Pi whose direction has the slope 2 Xi. 
This line is by definition tangent to the graph at Pj , or the 
graph is tangent to the line at Pi ; the point Pi is called the 
point of contact. Combining the above results we have : 

The graph of the function y = x"^ is a continuous curve, above 
the X-axis, symmetrical with respect to the y-axis, and passing 
through the origin. At any point Pi {xi , i/i) on the curve, the 
straight line ivith slope 2 Xi passing through this point is tangent 
to the curve. 

73. Further Observations regarding the Function y = x\ 

The preceding result tells us that when x= 1, the slope of the 
tangent is 2. Reference to Fig. 45 will verify this result for 

* This follows also directly from the formula m = {yz— 2/i)/(x« — ^i)- 



104 



MATHEMATICAL ANALYSIS 



[IV; § 73 



the straight line there drawn, since this line has the slope 2. 
In Fig. 47 we have reproduced Fig. 43 except that we have 
replaced the several points plotted in the earlier figure by a 
continuous curve and have drawn the tangent at the point 




P(l, 1). Knowing that the slope of the tangent is 2, we can 
easily construct the tangent. Starting from P we lay off any 
convenient distance PM to the right and then lay off double 
this distance MQ upward. The line PQ is then the required 
tangent. A similar process leads to the construction of the 
tangent at any other point of the curve. 

From the fact that the slope of the tangent at any point on 
the curve whose abscissa is Xi is 2 a^i, we see that as Xi increases 
numerically the slope increases numerically, that is, the curve 
becomes steeper and steeper the farther we go from the origin. 
Also the slope is positive when x^ is positive and negative when 
Xi is negative. This means that going from left to right the 
curve slopes downward at the left of the origin, and upward at 
the right of the origin. When a; = 0, the slope is zero, that is 
to say, the tangent is parallel to the a;-axis (here it coincides 
with the aj-axis). 



IV, § 73] 



QUADRATIC FUNCTIONS 



105 



Hitherto in our drawings we have chosen the unit on the 
t/-axis to be equal to that on the cc-axis. This renders it im- 
possible to draw the graph of the function y = x"^ for large 
values of a;, without making it of unwieldy size. However 







t I 


t 7 


. _ 


- t 7 


I } ^ 


^ ^ ^ 


t Z 


r / 


^z~ 


- z^i- 


// en 


J -U 


t ^ ^ 


"I ::^^: : : : 


A-- -«-- i' 


X ^ 


J ir 


:: ~A~r- : : : 


^<^ 


:- = r'V.. _ . 



23456789 10 

Fig. 48 



nothing prevents us from choosing the unit on the t/-axis 
smaller than that on the ic-axis, and in Fig. 48 we have chosen 
it one tenth as large. A tabular representation is as follows : 



X 


±1 


±2 


±3 


±4 


±5 


±6 


±7 


±8 


±9 


±10 


y 


1 


4 


9 


16 


25 


36 


49 


64 


81 


100 



In this case the slopes of the tangents are, respectively, 

±2, ±4, ±6, ±8, ±10, ±12, ±14, ±16, ±18, ±20. 
We have drawn the tangent at the point for which x = 5, and 
have drawn the graph only for positive values of x. 

Example. Find the equation of the tangent to the graph of 
2/ = a;2 at the point (3, 9). 

The slope of the tangent at the point (x^ , 2/1) is 2 a?] . There- 
fore at (3, 9) the slope is 6. The equation of the tangent is, 
therefore, y — 9 = 6{x — S) oy y = 6x — 9. 



106 MATHEMATICAL ANALYSIS [IV, § 73 

EXERCISES 

1. Discuss the functions y = — x^ ; y = 2x'^ \ y =— 2x^. 

2. Construct for the point (2, 4) of the function y = x^ n. figure anal- 
ogous to Fig. 46. (Use a table of squares.) 

3. Use the adjoining figure to give a geometric interpretation of the 
X Ax equation Ay = 2 xiAx i- Ax"^. The function y —x'^ is 

Ax ^^^® interpreted as the area of the square whose side 

is X. 



f////////M 


% 




% 



Ax 



4. If in the function y = x^ we take x = 3, how 
small must Ax be taken in order that Ay shall be 
numerically less than 0.01 ? if we take x = 15 ? Is 

the difference between these two results to be expected in view of the 

nature of the graph ? 

5. Draw the tangents to the curve 2/=x2 at the points for which x=0, 
± i, ± 1, ± -I, ± 2. 

6. If X is the radius of a circle and y is its area, prove that the 
change ratio Ay / Ax approaches the length of the circle as Ax approaches 
zero. 

7. Find the equations of the tangents to the curve ?/ = x^ at the 
following points : (1, 1); (2,4); (-1, 1); (-2, 4). Construct the 
tangents at these points. 

8. The line perpendicular to the tangent at the point of contact is 
called the normal to the curve at this point. Find the equations of the 
normals to y = x^ at the points (1, 1) ; (2, 4) ; (— 1, 1) ; (— 2, 4). Con- 
struct each normal making use of its slope. 

Ans. For the point (1,1): x + 2y-S =0.* 

9. Find the slope of the tangent to y = 3 x^ at the point whose abscissa 
is xi. What is the value of this slope at the point (1, 3) ? 

10. Find the equations of the tangent and the normal (see Ex. 8) to 
y = 3x2 at the points (3, 27) ; (- 2, 12). 

11. Find the points where the slope of the curve y = x^ has the values 
- 1 ; 2 ; 10. 

12. 1 cu. ft. of water weighs 66.4 lb. What must be the diameter x 
of a cylindrical can such that 1 in. of water contained in it will weigh 
y oz. ? Plot the graph and find x when y = 50. Find y when x = 8. 

* Assuming the units on the axis to be equal. 



IV, § 75] QUADRATIC FUNCTIONS 107 

74. The General Quadratic Function j/ = ax^ + 6x + c. 

We may now dispose of the general case. Let 

y = ax^ -\- bx + c {a =^ 0) 

be any quadratic function (in the case y = x^, a was 1, while 
b and c were 0). Let x increase from the value Xi to the 
value Xi -f Ax, and suppose that this change in the value of x 
changes the value of the function from 2/1 to yi -f Ay. We desire 
to calculate the value of Ay and of the change ratio Ay /Ax. 
We have 

2/1 -f Ay = a(x^ + Axy-\- b(xi + Ax) + c, 

and 

2/1 = a^i^ + b^i + c. 
Subtracting, we obtain 

(2) A?/ = (2axi-\-b -\- a Ax) Ax, 
and 

(3) ^ = 2aXi-\-b-{-aAx (ifAa;^0). 
Aa; 

Equation (2) shows that Ay can be made numerically as small 
as we please, by choosing Ax near enough to 0. Hence we may 
say: 

Every function of the form y = ax"^ -\-bx+ c is continuous. 

Equation (3) shows that the change ratio Ay /Ax approaches 
as a limit the value 2axi + bsiS Ax approaches 0. Hence we 
may say: 

The slope of the tangent to the curve y = ax^ -\- bx -{- c at the 
point whose abscissa is Xi is equal to 2aXi 4- b. 

75. General Properties of the Function ax^ + 6x 4- c. The 

discussion in the preceding section and the exercises have 
furnished us with some information regarding some special 
functions of the form aa^ -{-bx -\- c. 



108 MATHEMATICAL ANALYSIS [IV, § 75 

It will now be shown that whenever the term in ar* is posi- 
tive {i.e. a is positive) the graph of the function is an inverted 





a >o a < o 

Fig. 49 Fig. 50 

arch as in Fig. 49 and that whenever the term in x^ is negative 
{i.e. a is negative) the graph is an arch like the one in Fig. 50. 
To prove this we need only consider the slope of the tangent 
to the curve as the point of contact moves along the curve. 
We have just seen that the slope of the tangent is given by the 
formula m — 2axi + & at the point whose abscissa is Xi . There 
is just one point on the curve for which this slope is zero, viz. 
the point whose abscissa is 

^ a 
Now let us write the slope m of the tangent in the form 

The number in the parenthesis, i.e., Xi -f 5/(2 a), is positive 
when Xi> — 6/(2 a) and negative when Xi < — 6/(2 a). Geomet- 
rically this means that this parenthesis represents a positive 
number for points to the right of the straight line x= —6/(2 a) 
and a negative number for points to the left of this straight line. 

Case 1 : a > 0. If a is positive, the slope m is positive for 
points to the right of the line x = — 6/(2 a) and negative for 
points to the left of this line. 

In other words, for all points of the graph to the left of the 
line x = — b/(2 a) the tangent slopes downward (as we go from 
left to right) and for all points to the right of this line the 



IV, § 76] 



QUADRATIC FUNCTIONS 



109 



hi/>o 



Fig. 51 



tangent slopes upward. The point for which a: = — 6/(2 a) has 
its tangent parallel to the aj-axis. This point is called the 
minimum point of the graph (Fig. 51). 

Case 2 : a < 0. Suppose on the other 
hand that a is negative. The slope m 
is then negative when Xi + &/(2 a) is 
positive and positive when x^ + 6/(2 a) is 
negative. The slope is therefore positive 
when Xi< — 5/(2 a) and negative when 
Xi> — b/(2 a). At the single point for 
which X — — (6/2 a) the tangent is parallel 
to the aj-axis. This point is called the maximum point of 

the graph (Fig. 52). 

When x = — 6/(2 a) the function y = ax^ -\- 

bx -\- c has a minimum value if a > and a 

maximum value if a < 0. 

The curve represented by the function 

y = ax^ 4- 6a; + c is symmetrical ivith respect 

to the line x = —b/(2 a). 

The proof is left as an exercise. 

Hint. Show that the points which have abscissas — 6/(2 a) + h and 
— 6/(2 a) — h have the same ordinate. 




Fia. 52 



76. Definitions. 

of the form 



The curve represented by an equation 
y = ax"^ -{- bx -{- c 



is called a parabola. The lowest (or highest) point on this 
curve, i.e. the point for . which x = — 6/(2 a), is called the 
vertex. The straight line through the vertex and per- 
pendicular to the' tangent at the vertex is called the axis 
of the curve. The parabola is symmetrical with respect to 
its axis. 



110 



MATHEMATICAL ANALYSIS 



[IV, § 77 



77. to draw the Graph of a Parabola y=ax'^-\~bx-^c. 

The preceding discussion enables us to draw the graph of a 
quadratic function without plotting many points. 

Example 1. Sketch the graph of y = 2 x^ — 6 x -\- 5. 

The slope of the tangent at {x^, y^ is, by § 74:^ m = 4:Xi — 6. 

The vertex of the curve is the point for which 4 ajj — 6 = 0, i.e. 

the point for which x^ = 3/2 ; the corresponding value of y is 

1/2 and the vertex is therefore the point (3/2, 1/2). This 

point is the minimum point of the 
curve. We plot this vertex V, draw 
the horizontal tangent at this point 
and the vertical axis. We desire a 
few more points and their tangents 
on each side of the axis and then 
we can draw the curve. For ex- 
ample, we have 



Fig. 53 




X 


y 


m 


1 


1 


-2 


2 


1 


2 





5 


-6 



Example 2. Sketch the graph 
of 2/ = — aJ" + 4 a; -f- 5. 

The slope of the tangent at 
{xi , 2/i) is m = — 2 a^i -f 4. The 
vertex of the curve is at the point 
for which — 2 a^i + 4 = 0, i.e. for 
which Xi = 2. The corresponding 
value of 2/i is 9. Therefore the 
vertex, which is the maximum 
point of the graph, is at (2, .9). 
The graph is given in Fig. 54. 




2 4 \ 6 
Fig. 54 



IV, § 77] QUADRATIC FUNCTIONS 111 

EXERCISES 

1. Tell which of the following functions have a maximum and which 
have a minimum value. Find this value in each case and the correspond- 
ing value of X. 

(a) 2x^ + Sx-9. 

Ans. Minimum value : — 17, when x =— 2. 
(&) 3 x2 -f 8 X - 6. 
(c) -6x^-\-l0x- 12. 
((?) 3 a;2 + 6 X - 7. 
(e) -x^ + 1. 

2. Find the coordinates of the vertex and the equation of the axis of 
each of the following parabolas. Sketch the curves. 

(a) 2/ = 2 ic2 + 5 X + 3. 
(6) 2^ = 3 x2 + 9 X - 6. 

(c) y = - 5 x2 + 10 X - 12. 

Ans. F= (1, — 7) ; axis, x=l. 

(d) ?/ = 3 x2 + 6 X - 7. 

(e) y=-x2+l. 

3. The area of a certain rectangle in terms of the length of its side 
X is ^ = X (100 — 2 x). Find x so that this area shall be a maximum. 

4. A point moves on a straight line so that its distance s from a fixed 
point O on the line at any time t is given by one of the equations below. 
Draw the (s, t) graph and in each case show that the variable point 
reaches, on one side of O, a maximum absolute distance from 0. Find 
this maximum distance. Does this maximum absolute distance correspond 
to a maximum or a minimum value of s ? 

(a) s = f2_4« + 3. * 

(6) s = 2fi-St + 10. 
(c) s = 3 + 6 « - 4 «2. 

5. Find the equations of the tangent and the normal* to the curve 
y = x2 — 3 X + 1 at the point (1, - 1). Ans. y --x; y = x-2. 

6. Find the equations of the tangent and the normal * to the curve 
y = — 2 x2 + 3 X - 1 at the point (1, 0) . 

7. Find the equations of the tangent and the normal* to the curve 
?/ = — 2 x2 + 4 X — 1 at the maximum point. Ans. y = I ; x = 1. 

8. Find the equations of the tangent and the normal* to the curve 
y = 3x2 — 6x+lat its vertex. 

* See Ex. 8, p. 108. 



112 



MATHEMATICAL ANALYSIS 



[IV, § 78 




78. The Graph of y - k = a{x- Kf. The face that the 
graphs of functions of the form y = ax- -f- 6a; 4- c, all have the 
same general shajje but are differently located with respect 
to the coordinate axes suggests that many of these graphs 
may consist of curves, which might be brought into coin- 
cidence by a suitable motion in 
the plane. That this is indeed 
the case results from the follow- 
ing considerations, which lead to 
a general principle of far-reach- 
ing importance. 

Suppose the graph of the equa- 
tion y = ax^ is moved parallel to 
itself through a distance and 
direction which carries the point to the point Q {h, k). 
What will be the equation between the x and y of any point 
P on the curve in its new position, the axes of coordinates 
remaining in their original positon ? This question is readily 
answered. Let P' be the position of P before it was moved. 
The equation y — ax^ then tells us that M^P — a • OM'^ for 
every position of P' on the curve in its old position. After the 
motion, the directed segments OM' and 3PP' become respec- 
tively the (Erected segments QR and EP. Hence, for every 
point P on the curve in its new position we have 

(4) RP=a'QR\ 

If the coordinates of P are (x, y) we have x = Oil/, y = MP 

and 

QR = x-h, RP=:y-k. 

Therefore, by (4), the curve in its new position is the graph 

of the equation 

(5) y — k = a ' (x — hy. 

While we have applied these considerations to the function 



IV, § 79] QUADRATIC FUNCTIONS 113 

y = ax^, the reasoning is general ; consequently we may formu- 
late the following principle : 

GEJfEEAL Principle. If in any equation between x and y ice 
replace x by x — h and y by y — Jc, the graph of the neiu equation 
is obtained from the graph of the origiyial equation by moving the 
latter graph parallel to itself in such a ivay that the point moves 
to the point (Ji, k). 

We shall have occasion to apply this principle often in the 
future. 

79. Transformation by Completing the Square. At present 
we may use the principle just stated to prove that the parabolas 
y = ax"^ -{-bx-^- c and y = ax^ are congruent curves. 

This follows at once from the preceding general principle, if 
we prove that the equation 

(6) y = ax^ -\-bx-\-c 
can be written in the form 

(7) y-Jc= a{x - hy. 
To do this we write (6) as follows : 



2, = a(^ + ^+ )+o, 



and then complete the square on the terms in the parentheses by 

adding the term If' / (4 a^). In order to leave the value of y 

unchanged we must also subtract a y. b^ / (4a2) = b^ /{4^a) from 

the expression. This gives 

rrr,^ f . ,bx , b^ \ , ¥ 

(7') y = a\x'^ + —-' ' • ~ 



4^^;"^' 4a' 



or 



= a[x-{- 



=) 



^ ' 4a V 2 

This is of the form (7) for the values 



114 MATHEMATICAL ANALYSIS [IV, § 79 

2 a 4tt 

The operation just performed is called the transformation by 

completing the square. It is found serviceable in a variety 

of situations. It may be used to advantage in connection with 

numerical examples. 

Example. Discuss the graph of y = — 2 x^ + S x— 9. 
We first write 

y=-2(a:2-4x+ )-9; 

and tiien 

?/=-2(:c2-4.x + 4)-9 + 8 
or 

2/ + 1=- 2 (a; -2)2. 

The graph is then obtained from the graph ot y =—2x^ by moving the 
latter parallel to itself so that its vertex moves to the point (2, —1). 

EXERCISES 

1. By reducing to the form y — k = a {x — h)^, discuss the graphs of 
each of the following functions. 

(a) y = 2x'^+12x-\-2. (d) y = 2x'^ - 1 x + S. 

(6) 1/ =4x2 + 6a:- 9. (e) y =- Ax"- + 1 x +2. 

(c) ?/ = - 3 x2 + 9 X + 10. (/) y = - 3 x2 - 8 X + 10. 

2. Show that the equation of the straight line y — yi = m (x — xi) may 
be derived from the equation y = mx by the general principle of § 78. 

3. The results of § 79 furnish a proof of the fact previously derived, 
that the vertex of the parabola y = ax2 + 6x + c is at the point for which 
x=— &/(2a). Explain. 

4. Equation (7') proves that if a > 0, the value x = — 6 / (2 a) gives the 
minimum value to y ; also that if a<0, the value x=— 6/(2a) gives 
the maximum value to y. Explain without using the graph. 

Write the following equations in the form a (x — hy + 6 (y — ^')2 = c, 
where a, b, c, h, and k are constants. 

6. x2-4x + 2?/2-8 2/ = 2. 8. x^ + y^-iy = 2. 
Ans. (x-2)2 + 2(2/-2)2 = 14. 9. a;2-8x + y2 = o. 

'6. -2x2 + 4x+?/2_4y_3_0. 10. 3x2-4x-y2 +2 =0. 

7. 4 x2 - 4 X + 2 y2 _ 3 y ^ 1 _ 0. 



IV, § 80] 



QUADRATIC FUNCTIONS 



115 



II. APPLICATIONS OF QUADRATIC FUNCTIONS 

80. Maxima and Minima. We have seen that a quadratic 
function ax^ -\-bx + c has either a maximum or a minimum 
value according as a is negative or positive. Numerous appli- 
cations involve the problem of finding this maximum or mini- 
mum value and the corresponding value of a;, as the following 
examples show. 

Example 1. A rectangular piece of land is to be fenced in and a 
straight wall already built is available for one side of the rectangle. 
What should be the dimensions of the rectangle in order that a given 
amount of fencing will inclose the greatest area ? 

Before beginning the solution proper we should note carefully the sig- 
nificance of the problem. The length of the fence being given, we may 
use it to inclose rectangles of a variety of shapes, as indicated by the 
dotted lines in Fig. 56. Some rectangle whose shape is between those 
indicated will inclose the maximum area. |.^,,^,,,,,,,,^^^^^^^^^ 



1 

X 



I — I 
Fig. 50 



To determine this shape is our problem. 

To do this, it is necessary to express the | ' y j 

area (the quantity we wish a maximum) 

as a function of one variable. 

Solution : Let the dimensions of the 
rectangle be x and y and suppose the given length of fencing is L. We 
then have 
(8) 2x + y = L. 

The area inclosed is J. = xy, which from (8) becomes 

A = x(iL-2x)=Lx-2 x^. 

Plotting this function, we have the parabola in Fig. 57. We desire to 
find the value of x corresponding to the vertex 
V of this parabola, for this gives the greatest 
value to A. The slope m of the tangent is 
given by the equation m = L — ix, and this 
is zero (tangent horizontal at V) when x = \ L. 
ZZ^ For this value oi x, y = ^ L. The maximum 
area is therefore obtained when the width is 
one half of the length. The maximum area 
F.G. 57 is i L^ square units. 




116 



MATHEMATICAL ANALYSIS 



[IV, § 80 



Example 2. Three streets intersect so as to inclose a triangular lot 
ABC. The frontage of the lot on BC is 180 ft. and the point A is 90 ft. 

back of BC. A rectangular 
building is to be constructed 
on this lot so as to face BC. 
What are the dimensions of 
the ground plan which will 
give the maximum floor 
area? 

In Fig. 58 we have drawn 
the lot ABC and have indi- 
cated by dotted lines two 
extreme plans. The ground 
plan sought must be somewhere between these two extremes. To deter- 
mine its dimensions we proceed as follows : 

Let X and y be the length of the sides of the ground plan. The floor 
area (neglecting the thickness of the walls) is 




(9) 



A = xy. 



In order to express A, for which we seek a maximum, in terms of x 
alone, we now proceed to express y in terms of x. The triangles ABC 
and ^ilifiV" are similar. (Why?) Hence we have 



This gives 



whence 

(10) 

From (9) and (10) we obtain 



MN^^LA 
BC DA 

x 90- 



(Why ?) 



180 90 ' 
y = - i a; -H 90. 
A = 90 X - I x^. 



This expresses the floor .area as a function of the side x. The slope of 
the tangent to the graph is given by 

mz=90 - X 



and this slope is zero when x = 90, which in turn gives (by (10)) y = 45, 
and therefore A = 4050. The maximum area is then 4050 sq. ft. and this 
is obtained by making the building 90 ft. long and 45 ft. deep. 
Draw the graph of the function A = 90x-^ ^ x'^. 



IV, §80] QUADRATIC FUNCTIONS 117 

We may note that in both of these examples, the function of 
which the maximum was sought was obtained as a function of 
two variables. The conditions of the problem, however, made 
it possible to express one of these variables in terms of the 
other and thus to obtain the desired function as a quadratic 
function of one variable, whereupon the solution was readily 
effected. The difficulty in this type of problem is usually in 
connection with the elimination of all but one of the variables. 
To solve such a problem it is necessary to keep in mind the 
following steps. 

(1) Decide, and express in words, of what function a maxi- 
mum or a minimum value is to be found. 

(2) Express this function algebraically. 

(3) If this expression contains more than one variable, use 
the conditions of the problem to find a relation or relations 
connecting these variables. 

(4) By means of the relation or relations found, eliminate 
all but one of the variables from the function of which a maxi- 
mum or minimum value is sought. 

(5) Proceed with the algebraic computation. 

EXERCISES 

1. The number 100 is separated into two parts such that the product 
of the parts is a maximum. Find the parts and the corresponding 
product. -^ns. 50, 50, 2500. 

Is it possible to separate 100 into two parts such that the product of the 
corresponding parts is a minimum ? Explain. 

2. Prove that the rectangle of given perimeter which has the maxi- 
mum area is a square. 

3. Find the greatest rectangular area that can be inclosed by 100 yd. 
of fence. 

4. Separate 20 into two parts such that the sum of their squares will 
be a minimum. 



118 



MATHEMATICAL ANALYSIS 



[IV, § 80 




5. A man desires to build a shed against the 
back of his house, the ground plan to be a rec- 
tangle. The roof is to be 1 ft. higher in the 
back than in the front (see the adjoined figure). 
He has on hand enough siding to cover 253 sq. ft. 
Allowing 18 sq. ft. for a door and assuming that 
the height from the ground to 
the lowest part of the roof is 8 
ft., what should be the dimen- 
sions of the ground plan in 
order to get the greatest floor area ? 

6. An underground conduit is to be built, the 
cross section of which is to have the shape of a rec- 
tangle surmounted by a semicircle. If the cost of 
the masonry is proportional to the perimeter, and if 
the perimeter is 30 ft., what should be the dimensions of the cross section 
in order that the conduit will have a maximum capacity ? 

7. The same problem as in Ex. 6 with the perimeter of the cross section 
given as a ft. 

8. Determine the greatest rectangle that can be inscribed in a given 
acute angled triangle whose base is 2 5 and whose altitude is 2 a. 

*9. In the corner of a field bounded by two perpendicular roads a 
spring is situated 8 chains from one road and 6 chains from the other. 
How should a straight path be run by this spring and across the corner so 
as to cut off as little of the field as possible ? 

Ans. 12 and 16 chains from the corner. 




81. Table of Squares. We have stated that the more 
important functions have been tabulated (§ 28). The function 
x^ is one of these. Tables of squares are very helpful in 
shortening computation. A comparatively rapid method of 
constructing such a table is given in Ex. 2 below. Here we 
may make use of our knowledge of the function x^ to see that 
for a sufhciently small interval in such a table, we are justified 
in using linear interpolation (§ 56). Indeed we have seen that 

* The function whose minimum is sought is not in this case quadratic 
An approximate solution may be obtained graphically. The solution may be 
computed by finding the slope of the graph from the definition of slope. 



IV, §81] QUADRATIC FUNCTIONS 119 

in a sufficiently small neighborhood of any point on the graph 
of y = x^, the graph differs as little as we please from a straight 
2 line. (See Fig. 45.) For example, if in the second 
table on p. 100 we confine ourselves to only three-place 



.94 .884 accuracy, we find that the successive differences in 
.95 .903 , p . , . -, , 

g22 the function are almost proportional to the corre- 

.97 .941 sponding differences in the variable. We give in 
.960 the adjoined table an extract from the table men- 
tioned. From this table we may conclude that 

(.953)2 =.909. 

This result is accurate only to the third decimal place. 

EXERCISES 

1. Find by interpolation from the above table the following : 

(.954)2; (.981)2; (9.66)2; (9.89)2. 

2. Compute by actual multiplication the squares of all the integers 
from 31 to 40. This method of computing a table of squares becomes very 

laborious. Write the results obtained from 
your computation in a column, and write op- 
posite each pair of successive squares their 
difference as shown in the adjoined beginning 
of such a table. These differences are called 
the flrst differences of the table. Do you ob- 
serve any regularity in the formation of these 
differences? Prove in general the law here 
suggested. 

[Hint. Consider the difference between A;2 and (k + 1)^.] 
Use this law to construct a table of squares from 41 to 100. 

3. If the successive differences of the Jirst differences are formed, we 
obtain the so-called second differences. Prove that in a table of squares 
of successive integers the successive second differences are all equal to 2. 
The first differences, therefore, have the character of a linear function. 
Hence show how to compute the exact value of (.S2.6)2 from the value of 
(32)2 and (33)2^ This process is known as quadratic interpolation. 



X 


a:2 


Difference 


31 
32 
33 


961 
1024 
1089 


63 
65 


34 






35 






36 







120 MATHEMATICAL ANALYSIS [IV, § 82 

III. QUADRATIC EQUATIONS 

82. Definitions. An equation of the form ax^ -\- bx + c = 0^ 
where a, b, and c are constants and a =^ 0, is called a quadratic 
equation. 

A value of x which when substituted in the equation 
ax"^ -\- bx -{- c = makes both members identical is called a root 

Example 1. ^s 3 a root of the equation 2x^— 5x-\-6 = 0? 
Substituting 3 for x, we find 2.3- - 5.3 +Q = 9 and not 0. Therefore 
3 is not a root. 

Example 2. Determine k so that one root of 2 kx- — 3 ar + 5 = shall 
be 1. 

Since 1 is to be a root, we have 2 A: — 3 -|- 6 = 0, or k =— 1. The 
e( [nation then becomes — 2 ic- — 3 a; + 5 = 0. 

83. The Roots of ax^-^bx-\-c =0. It follows from § 79 
that the equation ax^ -f- te + c = may be written in the form 

5 \2 52 _ 4 ofc 



V 2 ay 



4a 



provided a ^ 0. Dividing by a and solving for (x + 6/(2 a)), 
we have ^ ^^_^ 

1^ + 2^>=^U? 



ac 



or 



6 , V62-4ac 
2a 2a 



hence 
(11) 



b ± V62-4ac 
2a 



We have shown up to this point that if ax^ -{-bx-\- c has the 
value 0, then x must have one of the values given in equa- 
tion (11). We need still to prove the converse : If 



— bH-Vb2 — 4flc — 6 — V62 — 4ac 

X = X_r or x = , 

2a ' 2a 



IV, § 82] QUADRATIC FUNCTIONS 121 

then ax^ -\-hx-\- c will have the value 0. This can be done by- 
substituting the values of x in turn in the expression ax^-^-hx-^-c 
and simplifying the resulting expressions.''^ 

The last part of this proof is essential. We know that the 
converse of a true theorem may be false. t The first part of 
our discussion proved that no other values of x than those 
given by (11) will satisfy the equation aoi^ -\-bx-\- c = 0, but it 
did not prove that either of these values does satisfy the given 
equation. 

Equations (11) maybe used as a formula for solving a quadratic 
equation. Thus, solving 

2 aj2 _ 5 aj _ 13 = 

where a = 2, 6 = — 5, c = — 13, we have 



or 



5±V25-4(2)(-13) 
x_ 4 



5 ± Vl29 



Solution by Factoring. If the factors of a quadratic 
equation may be found readily, one may proceed as in the 
following example. 

Example. Solve x"^ - Sx -\- 2 =0. 
This equation may be written in the form 
(a;-2)(x-l) = 0. 



Therefore, 


x-2 = 


or 


X- 


-1 = 0, 


i.e. 


x = 2 


or 




x=l. 


Why? See § 48. 











* The converse can be proved at present only if &2 _ 4 ac is not negative. 
Why ? 

t Thus the converse of the true statement, " A horse is an animal," would 
be the false statement, " An animal is a horse." 



122 MATHEMATICAL ANALYSIS [IV, § 83 

EXERCISES 

Determine whether the roots of the following equations are as stated. 

1. x2_ 5a;4-6 =0 ;2, 3. 4. 2^2 _ 5^ + 3 = O; 1, - 2. 

2. a;2 + 5 x - 6 = ; 1, 2. 5. a;2 - 7 = ; V7, - V7. 

3. x2 - 12 a: + 30 = ; 5, 6. 6. 7 a:2 _ 2 x + 51 := ; 0, 1. 

In the following equations determine k so that the number beside the 
equation is a root. Find the other root. 

7. a;2 + 2 Ax - 5 = ; 1. 8. A:x2 - 6 x + A;2 - 1 = ; 0. 

Ans. A; = 2 ; other root = — 5. 9. kx'^ — 6 A:x + 11 = fc ; 2. 

Solve the following equations by means of the formula and also by 
completing the square : 

10. (ax + &)2 = 6 X. 15. sx2 + to - p = 0. 

11. (x-5)(7x-3) = 12. jg x2 (3 x + 2)2 ^.^ 
12 y + 5 y^ - 5 ^ g . ■ 4 1 

7 3 * 17. 3(5x2 -10)+ 2x- 5=0. 

13. x2 -f A-x - c?x2 + /i = 0. 18. x2 + (p- g)x-i)g = 0. 

14. wP':r^ -\- m{n—p)x — mp = 0. 

Solve the following equations by factoring : 

19. x2 _ 8 x + 15 = 0. 22. 3 x2 - 17 X + 10 = 0. 

20. x2 - 14 X + 48 = 0. 23. • 5 X + 14 = x2. 

21. 12 - X - x2 = 0. 24. a&x2 + a2x + h'^x + a6 = 0. 

25. A cross-country squad ran 6 miles at a certain constant rate and 
then returned at a rate 5 miles less per hour. They were 50 minutes 
longer in returning than in going. At what rate did they run ? 

Ans. 9 miles per hour. 

26. When a single row of rivets is used to join together two boiler 
plates, the distance p between the centers of the rivets is given by the 
formula 

^ = 0.56^+(?, 

where t is the thickness of the plate and d is the diameter of a rivet hole in 
inches. In a certain make of boiler the rivets are 1 inch apart and the 
plate is \ inch thick. Find the diameter of the rivet holes. 

27. How high is a box that is 6 ft. long, 2 ft. wide, and has a diagonal 
8 ft. in length ? 



IV, § 84] 



QUADRATIC FUNCTIONS 



123 



28. The effective area JS" of a chimney is given by the formula 
E = A — O.GVA^ Avhere A is the measured area. Find the measured 
area when the effective area is 25 square feet. 

29. Two men can row 12 miles downstream and back again in 5 
hours. If the current is flowing at the rate of 1 mile per hour, how fast 
can the men row in still water ? 

30. Find the outer radius of a hollow spherical shell an inch thick 
whose volume is 76 tt/S cubic inches. 

[Hint. The volume of a sphere is 47rrY3.] 

84. Graphic Solution. Example. Solve x^ — 4.x + 3=0 
graphically. 

Let us plot the graphs oi y == x^, y = 4:X— 3 with reference 
to the same set of axes (Fig. 59). We see that the two graphs 
intersect in two points, the coordinates of 
which satisfy both equations. Therefore the 
abscissas of these points are values of x 
which make the right-hand members equal, 
i.e., for which 

0^2 = 4a; -3 
or 

a;2_ 4a; 4.3^:0. 

The roots are seen to be 1 and 3. 

If the line and the parabola were tangent, what would you 
say concerning the roots ? If the line and parabola do not 
meet, what would you say concerning the roots ? 
This problem may be solved graphically in 
an entirely different way. We will plot the 
curve y =zx^ — 4:X-{-3 (Fig. 60). The abscissas 
of the points where this graph meets the a?-axis 
are roots of the original equation. Why ? 
Describe the roots if the parabola touches 
the ic-axis. What would you say concerning the roots if the 
parabola did not meet the ic-axis ? 




Fig. 59 




Fig. 60 



124 MATHEMATICAL ANALYSIS [IV, § 85 

85. General Theorems. 1. If r is a root of the equation 
ax"^ 4- 6a; + c = 0, then x — r is a factor of ax"^ -{-hx -\- c. 
Dividing ax"^ + hx -[- chy x — r, we obtain : 

X — r\ ax^ -\-hx + c \ ax -\-{}j -\- ar) 
ax"^ — arx 

{b -h ar) X -\- c 

(b 4- ar) x — {b + ar) r 

c -{- br -\- ar"^' 
Therefore 

ax"^ -\-bx -\- c= [cix + {b -\- ar)']\x — r] + c + 5r + ar^. 

But, by hypothesis, r is a root ; therefore, ar"^ + br + c = 0\ 

hence 

ax^ -\-bx -{- G = \_ax -\-{b -\- ar)'][_x — ?-]. 

2. Prove that ifx — r is a factor of ax"^ -\-bx -\- c, then x = r 
is a root of ax^ -\- bx + c = 0. 

3. Prove that if the expression ax^ -\- bx -\- c is divided by 
X — r, the remainder is ar^ -\- br -\- c. 

The Discriminant of the Quadratic. In § 83 we saw 
that the roots of the equation ax^ -{- bx -\- c = are 

-6+V62-4ac ^^^ -b -V62_4(^c 



The expression under the radical, namely, &2 _ 4 qc^ is called 
the discriminant of the equation, because it enables us to dis- 
criminate as to the nature of the roots. From geometric con- 
siderations we know that a quadratic equation with real 
coefficients a, b, c may have either two real distinct roots, two 
real equal roots, or no real roots at all. The above formula 
enables us to see the same truth algebraically. 

If b^ — 4:ac = 0, we say that there are two real and equal 
roots, each being — 6/2 a. 

If 52 _ 4 fl(< -> 0, there are two real and unequal roots. 



IV, § 85] QUADRATIC FUNCTIONS 125 

If 62 _ 4 ac < 0, there are no real roots. The roots of such 
an equation are called imaginary or complex. The properties 
of such numbers will be discussed fully in Chap. XVIII. 

If the discriminant b^ — A ac is a perfect square and the 
coefficients a, b, c are rational numbers, then the roots are 
rational. 

By finding the value of the discriminant we may determine 
the nature of the roots of the quadratic without solving the 
equation. Thus, in the equation Sx'^-i-ix — 3 = 0, the dis- 
criminant is 52 and we conclude that the roots are real, un- 
equal and irrational. 

Eelation of Roots to Coefficients. Let the roots of the 
equation ax- -f- 6a; -f c = be denoted by /•, and ra . That is, let 

— 64-V62 — 4ac „ 1 —b — Vb"^ — 4 ac 



and ?'2 = 
2a 2a 



By addition we have 



-5.|-V 52_4qe-5-V6'^-4ac _ 2b_ b 
'*^ + '*^- "" 2a ~ 2a~ a 

By multiplication we have 



_ r(- b)-^¥ - 4 ac\\{- b) + ^¥ - 4 ac'] 
'"''''- 4a^ 

_ 6^^ — 6^ 4- 4 ac _ c 
~ 4a2 a 

Therefore, if we write the quadratic equation in the form 

x^-\-^x + -=0, 
a a 

the above results may be expressed as follows : 

In a quadratic equation in ivhicJi the coefficient of the x^ term 
is unity, (/) the sum of the roots is equal to the coefficient of x 
with the sign changed; (ii) the product of the roots is equal to 
the constant term. 



126 MATHEMATICAL ANALYSIS [IV, § 85 

EXERCISES 

Solve graphically (two ways) each of the following equations : 

1. 2x2 -|-5x- 3:^0. 3. 12-a; = a;2. 5. 4-x2 = 0. 

2. a;2-8x + 15=0. 4. 2x2-3x-5=0. 6. 4 + x2 = 0. 
Form the equations with the following roots : 

7. 4,-5. Ans. x2+x-20:=0. 9. 2+V5, 2-V5. 

8. V7, - V7. 10. c + 3 &, c - 3 b. 

11. What is the remainder when 3x2 — 2x + 5=0 jg divided by 
x-3? byx + 2? byx-1? by-x+1? [Hint : Use 3, § 85.] 

Determine, the nature of the roots of the following equations : 

12. 7x2_5x = 6. 14. 2 1/2 + 3 2/ + 24 = 0. 

13. 2x = 7-3x2. 15. 9x2 = 4x-5. 
Determine k so that the following equations shall have equal roots. 
[Hint : Place b'^ — iac equal to zero.] 

16. kx^ - 6 x + 3 = 0. Ans. k=S. 18. x^ + 2 {1 -^ k)x -{- k^ = 0. 

17. 3iK2-4^•x+ 2 = 0. 19. 2Arx2+(5A: + 2)x + 4^ + l=0. 

20. Determine the limits on k so that equations 16-19 shall have their 
roots real and unequal ; imaginary and unequal. 

21. If X is real, show that ~ must lie between and 1. 

a;2-5x + 9 11 

22. A party of students hired a coach for ^ 12, but three of the students 
failed to contribute towards the expense, whereupon each of the others 
had to pay 20 cents more. How many students were in the party ? 

23. Cox's formula for the flow of water in a long horizontal pipe con- 
nected with the bottom of a reservoir is 

Hd^ iv^ + ^^v-2 
' L 1200 

where H is the depth of the water in the reservoir in feet, d the diameter 
of the pipe in inches, L the length of the pipe in feet, and v the velocity 
of the water in feet per second. If a reservoir contains 49 ft. of water, 
find the velocity of the water in a 5-inch pipe that is 1000 ft. long. 

24. It takes two pipes 24 minutes to fill a certain reservoir. The larger 
pipe can fill it in 20 minutes less time than the smaller. How long does 
it take each pipe to fill the reservoir ? Ans. 60 min. ; 40 rain. 

25. Prove algebraically and geometrically that if h^ — ^ac<:iQ^ the 
value of the function ax^ + &x -f c is positive for all (real) values of x, 
if a > ; and negative for all (real) values of x, if a < 0. 



IV, § 86] QUADRATIC FUNCTIONS 127 

86. Equations involving Radicals. The method of solving 
problems involving radicals will be illustrated by some 
examples. 

Example 1. Solve Va; + ^ - 2 (a; — 1)= 0. 

Transposing the second term to the right-hand member gives 

V»T2 = 2(x- 1). 
Squaring, 

x-\-2=^4.x'^-Sx-\-4:, or 4a.'2 -9 a; -f- 2 = 0. 

Whence 

, a; = 2, or i. 

Do both these values satisfy the equation? 

We have shown that, if VicH-2 — 2 (ic — 1)=0, then x = 2 
or J. But we cannot conclude conversely, that if a; = 2 or \, 
then VxT2 - 2 (it- - 1) = 0. 

In fact, if we substitute the values of x found in the original 
equation, we find that a; = 2 is a root ; but a; = ^ is not. 

Example 2. Solve the equation Va; + 8 + Vx + 3 = 5 Vx. 

Squaring both sides, we find 

a; -f 8 + 2Va;2-f llaj-f 24 + x + 3 = 25a;, 
or 



2 Va;2+ 11 a; + 24 = 23 x - 11 ; 
whence squaring, collecting terms, dividing by 25, we have 

21 a;2 - 22 a; + 1 = ; 
therefore, a; = 1 or ^V- 
What are the roots ? 

EXERCISES 

Solve each of the following equations : 



1. Vic - 2 - 3 = 0. 4. - \/4 X - 3 - Vx + 1 = 1. 



2. >/x-2-(-3 = 0. 5. v'x+5+ Vx+10=v2x + 15. 



Ans. No roots. $. Vx4- b + Vx-\- a = V2x+a + b. 
3. Vx~+2-VxTT = - 1. 7. V2 X + 6 - Vx + 4 = Vx - 4. 

Ans. 2. 8. Vx+3 - V4 x + 1 = V2 - 8 x. 



128 MATHEMATICAL ANALYSIS [IV, § 86 

MISCELLANEOUS EXERCISES 

Determine the condition existing among a, 6, c so that the equation 
ax'^ + bx + c = shall have : 

1. One root double the other. Ans. 2b^ = 9 ac. 

2. The roots reciprocals of each other. Ans. a = c. 

3. One root three times the other. 

4. One root n times the other. 

5. One root zero. Ans. c = 0. 

6. One root equal to 1 ; 2 ; 3 ; n. 

7. The roots numerically equal but opposite in sign. A7is. 6 = 0. 

8. Find the area of the largest rectangle that can be inscribed in a 
triangle whose base is 20 inches and whose altitude is 15 inches, if one 
side of the rectangle is along the base of the triangle. 

9. Separate twenty into two parts such that the product of half of one 
part by a quarter of the other shall be a maximum. 

10. Solve the equation y^ — Sy"^ -^ \b = 0. [Hint : Let y^ = x.] 

11. Solve the equation fa; + - 1 + T^ + ^1 — 12 = 0. 

12. Solve the equation x^ + 8 x + 3 V^ + 8x4-2= 8. 

13. Solve the equation — ^^^ = — • 

x + l x-^ 12 

14. Find k so that the roots of {k + 2)x^ — 2 A;x + 1 = are equal. 

15. Without solving, determine the sum and product of the roots of each 
of the following equations : 

(a) 2.r2_7x-3 = 0. , (c) 4ic2 _ 3^: + 1 =0. 

(6) x2 - 4 X + 2 = 0. (d) 2 a;2 + .3 x + 4 = 0. 

16. Determine k so that the sum of the roots of the equation 
2x2+(A-- l)x+(3A: -7) = 0is4. Ans. k = -l. 

17. Determine k so that the product of the roots of the equation 
(2 A; - 1) x2 + (A; + 3) » + (A;2 - 2 A; + 1) = is 2. 



CHAPTER V 



THE CUBIC FUNCTION. THE FUNCTION a^ 



87. The General Cubic Function ax^ -f- bx"^ -\- ex -\- d. Hav- 
ing discussed in the last chapter the general quadratic function 
ax^ -{-bx -\- c, we now turn our attention to 
the general algebraic function of the third 
degree, i.e. the general cubic function. It 
is of the form 



OQ^ -\- bx^ -\- ex + dj a=^ 0. 



88. The Function ^. We begin with 
the consideration of the function y = a^. 

A brief tabular representation of this 
function is given below. 

We note that the values of x^ 
for negative values of x are the 
same in absolute value as those 
for the corresponding positive 
values, but negative. If the 
corresponding points are plotted 
with respect to a pair of rec- 
tangular axes, we obtain Fig. 61. 

The change Ay in y due to a 
change Ax in x is calculated as follows, where x^ and y^ are 
any pair of corresponding values of x and y : 

(1) y^ + Ay = x^ + 3 a;i2 . Aa; H- 3 x^^'^ + Aa^. 

K 129 



X 


a;3 





0.00 


.5 


0.12 


1.0 


1.00 


1.5 


3.36 


2.0 


8.00 


2.5 


15.62 


3.0 


27.00 



















^j^ 
































x 




























in 










































\ ' ' 


--3^'Alh 


--Jr-S^S--^ 



















































































































Fig. 61 



130 

Since 

this gives 
(2) 



MATHEMATICAL ANALYSIS 

2/1 = ^ly 



IV, § 88 



A?/ = (3 x^^ + 3 x^^x + Aa;2) Aa;. 

We can now conclude that as Aa; approaches zero, Ay also 
approaches zero ; i.e. the function is continuous for all values 
of X. From (2) we obtain 



(3) 



^ = 3a;i2-j-3a^iAa; + Aa-2 

Ao; 



(if Ax z^ 0). 



As Aaj (and, therefore, also A?/) approaches 0, this change ratio 
approaches 3 x^. The slope m of the graph at the point {x^ , y^ 
is, therefore. 



(4) m = ^x^\ 

This slope is positive for all values of Xy^ 
except a?! = 0. Why ? The function is 
therefore an increasing function for all 
values of x except a? = 0, i.e. at the origin, 
where the graph of the function is tangent 
to the X-axis. The graph is exhibited in 
Fig. 62, where we have drawn at certain 
points the tangents to the graph by means 
of (4) in order to insure greater accuracy. 














- i^ :: J::::: 


1 -L- .- 


± JL :: 






-"M : 3^::::: 


di T 






::^ : :t::::: 


«' , i 


'-% 


"r ' jL 






Jrf T '1- e> 9 7: '. 


:. ::::?_>:3 ^-4i5- 






/ -^ _. 


it > 




"L — ":::: :: 
















:: : ■-!«!: ::::: :: 











Fig. 62 



89. The Functions ax^ and a{x - hf + ft. 

From the results of the last article and the 

general principles previously established, we conclude that the 

graph of the function 

y = ay? 

is obtained from that of 2/ = a^ by stretching or contracting all 
the ordinates in the ratio | a | : 1, according as \a\ is greater 
than 1 or less than 1, and in case a is negative reversing the 



V, § 89] 



CUBIC FUNCTIONS 



131 



signs of all the ordinates (Fig. 63).* Explain the reason for 
this result. 

The function y == a{x ~ Tif -{- k may be written in the form 

y — 7c = a{x — hy. 

Its graph is accordingly (§ 78) obtained from that of ?/ = aa^ 
by sliding the latter graph through a distance and in a direc- 




y=a(z-h)3 + k 
Fig. 64 

tion represented by the motion from (0, 0) to (7 = {h, 7c). Ex- 
plain the reason for this (Fig. 64). 

The slope of y = aa^ at the point (xi, y{) is 3 axi^. The slope 
of a (a; - hy+ k at the point {x^, y{) is Sa{xi- hy. The proof 
of these statements is left as an exercise. 



EXERCISES 

1. From the graph of the function y — cc^, determine the volume of a 
cube whose edge is 0.5 in. ; 0.5 ft. ; 3 f t. ; 1.5 cm. 

2. Find the equation of the tangent and the normal to the curve y =.3^ 
at the point (2, 8) ; (_ 1, _ 1) ; (0, 0) ; (- 2, - 8). 

3. Draw each of the curves y= -x^, y=4i x^, y=S(x-iy, y=2(x-\-iy 

* For example, if the unit on the y-scale of Fig. 62 be doubled (i.e. made 
equal to the a;-scale) while the curve is left unaltered, the graph there given 
will be the graph of y = i cc^. 



132 MATHEMATICAL ANALYSIS [V, § 89 

4. Show that the slope oiy = — ax^ at the point (xi, 2/1) is — 3 axi^. 

6. Discuss the locus of y = — xK 

6. Discuss the locus of y =— ax^ if a is positive and greater than 1 ; 
less than 1. Show that the same curve will serve as the graph for all 
values of a > if the units on the axes are properly chosen. 

90. The Addition of a Term mx. Shearing Motion. If to 

an expression in x defining a function, a term of the form mx 
be added, the effect on the graph is readily described in terms 
of a type of motion that is important in mechanics. For ex- 
ample, let us take the function a^ and investigate the effect 
produced upon the graph by adding the 
term —Sx. The graphs of y = a^ and 
y=:—Sx are dravm in Fig. 65. The graph 
of y = a^ — Sx is then obtained by adding 
^ /" \-A=r?--5a the corresponding ordinates of the former 
/ ^ graphs. The addition of these two func- 

tions is obtained graphically by sliding the 
ordinate of each point on y = a^ vertically 
up or down until the base of that ordinate 
meets the graph of y =: — Sx. If we think of the ordinates 
of 2/ = 25^ as attached to the avaxis and constrained to remain 
vertical, the graph of y = a^ will become the graph of 
y = a^ — 3x if the a>axis is rotated about the origin until it 
coincides with the line y = — Sx. The resulting graph of 
2/ = 0? — 3 a; is, of course, to be interpreted as drawn with ref- 
erence to the original a>-axis. The motion just described, 
whereby y = 0^ is transformed into y = a^ — Sxj is called a 
shearing motion or a shear with respect to y = — Sx. 

In general, if the term mx is added to aa^, the graph of the 
function cta^ -{-mx is obtained by subjecting the graph of cue* 
to a shear with respect to the line y = mx. .If a and m have 
the same signs, the effect is in the direction of straightening 




V, § 90] 



CUBIC FUNCTIONS 



133 



the graph ; if a and m have different signs, the effect is in the 
direction of emphasizing the curvature. 

These effects can be produced by drawing the original figure 
on the edges of a pack of cards, or on the edges of a book, and 
then shifting the cards (or sheets of paper) as shown in Fig. 66. 




EXERCISES 

Draw the graph of the following functions, making use of the shear : 

1. y = Sx'^ + x. 5. y=:ic8 + x-l. 

2. ?/ = x2 + X. B. y=-x^ + x-^2. 

3. y=-x^-x. 7. y = x^-l. 

4. y = - 2 x3 + 4 x. 8. y = x2 - 4 X. 

9. Show that y = mx is the equation of the tangent to the curve 
= x^ 4- mx at the origin. 



134 



MATHEMATICAL ANALYSIS 



[V, § 91 



91. The Functions a(x - hf + m{x - h)-\- k and ax^ + bx^ 
-{- cx + d. We have seen that the graph of 



(6) 



y = ax^ -f mx 



has one of the following forms (Fig. 67) according to the signs 
of a and m. 

If such a graph is subjected to a parallel motion which 




Fig. 67 

carries the origin to the point {h, k), the equation of the graph 
in its new position is (§ 78) 

(6) y — k = a{x — hy -\-m{x— h), 
which when expanded takes the form 

(7) 2/ = «^' — 3 ahx^ + (3 ah^ -\-m)x— ah^ — mh -f k. 
This is of the general form 

(8) y =z ao? -^ hx^ -[- ex + d. 



V, § 93] CUBIC FUNCTIONS 135 

Moreover, it includes for all values of 7i, k, m all the equations 
of the general form (8). For (7) and (8) will be identical if 

(9) - 3 aA = 6, 3 a/i2 -(- m = c, - ah^ - mh -{- k = d. 

The first of these equations determines h(a^O) ; h being 
known, the second equation determines m; m and h being 
known, the third equation determines k. ■ We may conclude 
then that the graph of any function of the form (8) has one of 
the shapes given in Fig. 67, but with the origin moved to a 
point {h, k) given by the equations (9). 

In order to draw the graph of a function of the form (8) 
we could first transform (8) into the form (6) and then pro- 
ceed as in § 90. It is more expeditious, however, to proceed 
more directly by making use of the slope of the function (8) 
and our knowledge of what shapes may be expected. 

92. The Slope of y = ax^ + bx^ -\-cx + d. The change Ay 
in y due to a change Ax in the function 

y = aa^ -\- bx^ -{-ex -j- d, 

when a; = aji , is 

Ay = (3 axi^ -{- 2 bx^ -\- c -\- 3 ax^Ax +- bAx -f aAx'^) Ax. 

This equation shows that the graph is continuous. Why? 
When Ax approaches 0, the change ratio Ay /Ax approaches 
the slope m, by definition. This gives, 

93. To draw the Graph of y = aj^ -h bx^ -\- ex -{- d. We 

shall illustrate by means of two examples the method of draw- 
ing the graph of a cubic function. 



136 



MATHEMATICAL ANALYSIS 



[V, § 93 



Example 1. Draw the graph oi y = a^ + x'^ — x + 2. 
The slope m at the point (a;i, yi) is (§ 92) 

m = S Xi^ -\- 2 Xi — 1. 

We seek first the points (if such exist) at which the tangent 

is horizontal, i.e. where m = 0. The roots of the equation 

m = 0, viz. 

3a;i2 + 2a;] -1 = 

are Xi = — 1 and Xi = -|. The slope is therefore at the 
points (-1,3) and (^, If). 

We now compute a table of corresponding values of x, y, m 
for values of x on both sides of and between x — 1 and x = ^. 
Such a table and the corresponding figure are given below. 



X 


y 


m 


-3 
-2 


- 13 



20 

7 


-1 


3 








2 


-1 


1 


If 
3 



4 


2 


12 


15 



Fia. 68 



Example 2.* Draw the graph of y = — a^ — Sx^ — Sx-^-l. 
The slope at the point where x = Xi is 

(3a;i2 + 6aJi + 5). 



m 



Since the roots of the equation 3 iCi^ -f 6 cci + 5 = are im- 
aginary, the graph has no horizontal tangents and the slope m 
is negative at every point. We accordingly make a table of 
values and construct the graph (Fig. 69). 



V, § 94] 



CUBIC FUNCTIONS 



137 



X 


y 


m 


-8 


16 


-15 


-2 


7 


- 5 


-1 


4 


- 2 





1 


- 5 


1 


-8 


-14 




Fia. 69 

94. Maxima and Minima. We extend our definition of 

maximum and minimum given in § 75 as follows : 

A value of x for which a function stops increasing and 
begins to decrease is said to correspond to a maximum of the 
function ; a value of x for which the function stops decreasing 
and begins to increase is said to correspond to a minimum of 
the function. Thus in Ex. 1, § 93 the value x = —1 corre- 
sponds to the maximum 3 of the function ; the value a; = -J 
corresponds to the minimum ^ of the function.* 

EXERCISES 
Draw the following curves and locate in each case the maximum and 
minimum points if there are any : 

6. y = oc^ ■\- X + 1. 

7. y = oc^_. 

8. y = x^ — X. 

9. y =z x^ + 2 x"^ -h X. 



1. 


y = x^ + x^. 


2. 


v=f-5|!-a.., 


3. 


y^X^- — -2x-\-\' 


4. 


y = x^-x'^-5x+2. 



6. y = 2a;8 4-^-4x + l. 



10. y 



cfi -x^ -\-x—l. 



* Note that a maximum of a function does not mean the greatest value a 
function can assume. In Ex. 1, § 93, the value of the function is greater when 
x = 2 than when x = —l. It does mean a value of the function which is 
greater than the values in the immediate neighborhood. 



138 



MATHEMATICAL ANALYSIS 



[V, § 9; 



95. Geometric Problems in Maxima and Minima. The 
theory just explained has an important application in solving 
problems in maxima and minima, i.e. the determination of the 
largest or the smallest value a magnitude may have which 
satisfies certain given conditions. 

As we saw in § 80, the first step is to express the magnitude 
in question algebraically. If the resulting expression contains 
more than one variable, other conditions always will be given 
which will be sufficient to express all of the variables in terms 
of one of them. When the magnitude in question is expressed 
in terms of one variable, we can proceed as in § 92 to find any 
maximum or minimum values which there may be. 

Example 1. Find the greatest cylinder that can be cut 
from a given right circular cone, whose height is equal to the 
diameter of its base. 




Fig. 70 



Let li be the given height of the cone and x and y the un- 
known dimensions of the cylinder (Fig. 70). Then the volume 
V of the cylinder is equal to irx^y. But from similar triangles 



we have 



V, § 95] CUBIC FUNCTIONS 139 

Therefore, 



whence 
Now 



F= TTxXh -2x) = Trhx^ - 2 no^, 
m = 2 irhx — 6 ttx^. 



The roots of the equation m = are a; = and x =h/S. 
It is left as an exercise to draw the graph of the function 

V = Trhx" - 2 TTCl^ 

and show that the value x = h/3 corresponds to the maximum 
of the function, i.e. to y = h/S. Therefore the maximum 
volume of the cylinder is obtained when the altitude is equal 
to the radius of the base. The maximum volume is 7r^'/27 
or 12/27 of the volume of the cone. 

EXERCISES 

1. A square piece of tin, the length of whose side is a, has a small 
square cut from each corner and the sides are bent up to form a box. 
Determine the side of the square cut away so that the box shall have the 
maximum cubical contents, Ans. a/6. 

2. Assuming that the strength of a beam with rectangular cross section 
varies directly as the breadth and as the square of the depth, what are 
the dimensions of the strongest beam that can be sawed from a round log 
whose diameter is d. Ans. Depth = Vf d. 

3. Find the right circular cylinder of greatest volume that can be in- 
scribed in a right circular cone of altitude h and base radius r. 

Ans. Radius of the base of the cylinder equals f r. 

4. Equal squares are cut from each corner of a rectangular piece of 
tin 30 inches by 14 inches. Find the side of this square so that the re- 
maining piece of tin will form a box of maximum contents. 

5. Show that the maximum and minimum points on the curve 
y = x^ — ax + 6 (a > 0) are at equal distances from the y-axis. 

6. Find the maximum volume of a right cone with a given slant 
height L. 



140 



MATHEMATICAL ANALYSIS 



[V, §96 



96. The Power Function. The functions x" and l/a;», where 
n is any positive integer, are called power functions of x. The 
curves y = .t'» (Fig. 71) are known as parabolic, while the curves 
y = l/iC* (Fig. 72) are known as hyperbolic. 

The curves of the parabolic type possess the property that 
they all pass through the point (0, 0) and the point (1, 1). 



r 










Y 










T 


H 


" 






\ 






















4 


. ili/L 








\ 






















*f 


/ 


4- 


^ 








\ 


a 


















>. 










\ 


H 




-g 


>%- 


.- 










'■ 


Tr 




^ 




._, . 




V 


, V 














/ 










/ 


\ 






u 
> 

-8 


C 
























\/ 






\- 
























-> 


V 






f 




























\ 






.. 
















/ 


/ 










\ 




4) C*J 


















/ 












\ 




>' 


















/ 


















N 
















ill 




















\ 


















w 






































1^ 






































4i 




















\ 
















/ 


m 






















s, 












/ 


/ 


fit 




















[ 




s 








/ 


/// 


1 






















^ 




»^ 




A 


^ 


^ 


-> 


' 
















'V 




_![ 


/ 


^ 


=; 


Z' 


a 






/ 






j5 








X 




_j 


'/ 




/ 
































// 


' 


/ 


































f 


/ 




































/ 


J 
































'V 




^14. 


- 








PoLJv, 


J. 


i 


n, , 


rv' 


JS 


- 


- 


-tsi 


^-^ 


5 








V 


Iclk. 




"^ 














1 




?/=LrcZl__ 










y^ 


-%] 


[ ^ 


ti 


1 










, 




^i72r,3;£ 


)j"l 


)- 


- 


- 


5 








i 


or-p 




Ji 


■;3- 


i 






























^3 






































5 


































Q 




































J- 












1 

















Fig. 71 



The larger the value of n, the greater is the slope of the tan- 
-^ent at the point (1, 1). 

The curves of the hyperbolic type all pass through the 
point (1, 1). As X approaches 0, the corresponding value of y 
becomes infinite. At a; = the value of y is undefined. As x 
becomes infinite, the corresponding value oiy approaches 0. 



V, § 96] 



CUBIC FUNCTIONS 



141 













-y 




_1M. 


1 














































Jv J]J 
























111 
















^ 


•< ^ 




~\h 


,>^ 














^ 


il. 






V^l 
















^ 






y 


jLii-L 


















Hr 






SpL 


~^ 
















j 






^\ \ 11 


















T_^ 






VjLl 


"^ ::ix 










l_ 




/ 






J4l 


1 1 














1 






\\ 


















[7 






\\ 


\| 
















\t 






\J 


vl 
















1 








^ 














J 




^ 






V, 




■^ 










/ 










V'^ 












)^'> 


f 








r-lO^ 


\' 


■^& 


=^ 









li 









X y-. 


m^^ 




■^~' 


--= 


^~'*p 


CT 




Y) 


-- 






rrr 


gmijj,.^] 






s 


f 










1 




1 


"" "— 


"--L 


^^\ 
















_i_ 






-:k 


s\ 








T 


ly-perbo 


■ck 








^'S 


1 


















\ 


1 






1 




r/ /»# L 




















-x 


.^i?:._ 


















fo 


"has 


1,2, 

-tv/c 


^,10 










\ 








]yh 


' 1 X 










\ 






r. 


) Darts 










\ 
























^ 


























I 
















































\ 


-A 






J. 







Fig. 72 



EXERCISES 

1. Draw the curves y =x^\ y = x^ ; y = x^; y = x^. 

2. Draw the curves y = l/x; ?/ = l/x^ ; y = 1/x^ ; y = 1/ic*. 

3. Prove that the slope of the tangent at the point (1, 1) to the curve 
y = x2, is 2 ; to the carve y—x^ is 3 ; to the curve y=x^ is 4 ; lo the curve 
y = x^ is 5. 

4. Prove that for every even value of n, the parabolic curves ' ?/ = x" 
pass through the point (- 1, 1); and that for every odd value of n, they 
pass through the point (— 1, — 1). 

6. Prove that the function x^ is an increasing function for all values 
of x. 

6. Find the equation of the tangent and the normal to y = ar^ at the 
point (2, 32). 

7. Prove that the slope of the curve y = 1/x at the point (xi, yi) Is 
— 1/xi*. [The curve y — l/x is called a hyperbola.'] 



142 



MATHEMATICAL ANALYSIS [V, § 96 

ace* be obtained from the 



8. How can the graph of the function y 
graph of y = x" if a is positive ? negative ? 

9. Find the equation of the tangent and the normal to the curve 
y = l/a:2 at the point (2, \) . 

10. Prove that all hyperbolic curves lie within the shaded regions of 




the adjoining figure, while all parabolic curves lie in the regions left 
unshaded. 



CHAPTER VI 
THE TRIGONOMETRIC FUNCTIONS 

97. The functions we have discussed hitherto, namely, the 
functions of the form mx i-b, ax^ -^bx -\- c, ax^ + bx"^ + ex -\- dj 
have all been defined by means of explicit algebraic expres- 
sions. They are all examples of a very large class of functions 
known as algebraic functions. We now turn our attention to 
functions defined in an entirely different way. As we shall 
see, these functions depend on the size of an angle. They 
enable us to express completely the relations between the 
sides and the angles of a triangle, and they are of the 
greatest practical importance in surveying, engineering, and 
indeed in all branches of pure and applied mathematics. 

98. Directed and General Angles. In elenJentary geometry 
an angle is usually defined as the figure formed by two half- 
lines issuing from a point. However, it is often more serviceable 
to think of an angle as being generated 
by the rotation in a plane of a half-line 
OP about the point as a pivot, start- y^ \ 
ing from the initial position OA and / ^^-^"'^ \ 
ending at the terminal position OB (Fig. 
73). We then say that the line OP has 

generated the angle AOB. Similarly, if OP rotates from the 
initial position OB to the terminal position OA, then the angle 
BOA is said to be generated. Considerations similar to those 
regarding directed line segments (§ 6) lead us to regard one of 

143 



144 MATHEMATICAL ANALYSIS [VI, § 98 

the above directions of rotation as positive and the other as 
negative. It is of course quite immaterial which one of the 
two rotations we regard as positive, but we shall assume from 
now on, that counterclockwise rotation is 
positive and clockwise rotation is negative. 
Still another extension of the notion 
of angle is desirable. In elementary 
geometry no angle greater than 360° 
is considered and seldom one greater than 180°. But from the 
definition of an angle just given, we see that the revolving 
line OP may make any number of complete revolutions before 
coming to rest, and thus the angle generated may be of any 
magnitude. Angles generated in this way abound in practice 
and are known as angles of rotation.^ 

When the rotation generating an angle is to be indicated, it is 
customary to mark the angle by means of an arrow starting at 
the initial line and ending at the terminal line. Unless some 
such device is used, confusion is liable to result. In Fig. 75 



e^ 0^ Q 




30' 390" 750' lilO 

FiQ. 75 

angles of 30°, 390°, 750°, 1110° are drawn. If the angles were 
not marked one might take them all to be angles of 30°. 

99. Measurement of Angles. For the present, angles will be 
measured as in geometry, the degree (°) being the unit of measure. A 
complete revolution is 360°. The other units in this system are the 
minute ('), of which 60 make a degree, and the second ("), of which 60 
make a minute. This system of units is of great antiquity, having been 

* For example, the minute hand of a clock describes an angle of — 180® 
in 30 minutes, an angle of — 540° in 90 minutes, and an angle of —720° in 
120 minutes. 



VI, § 101] TRIGONOMETRIC FUNCTIONS 



145 



used by the Babylonians,* The considerations of the previous article then 
make it clear that any real number, positive or negative, may represent an 
angle, the absolute value of the number representing the magnitude of 
the angle, the sign representing the direction of rotation. 

100. Angles in the Four Quadrants. Consider the angle 
XOF = e, whose vertex O coincides with the origin O of a system of rec- 
tangular coordinates, and whose initial line OX coincides with the positive 



-p^Y-. -0^^ 




Fig. 76 

half of the x-axis (Fig. 76). The angle d is then said to be in the first, 
second, third, or fourth quadrant, according as its terminal line OP is in 
the first, second, third, or fourth quadrant. 

101. Addition and Subtraction of Directed Angles. The 

meaning to be attached to the sum of two directed angles is analogous to 
that for the sum of two directed 
line segments. Let a and h be 
two half-lines issuing from the 
same point and let (a&) repre- 
sent an angle obtained by rotat- 
ing a half-line from the position 
a to the position 6. Then if we 
have two angles (a&) and (6c) with the same vertex 0, the sum (a6)-|-(6c) 
of the angles is the angle represented by the rotation of a half-line from 
the position a to the position h and then rotating from the position 6 to the 
position c. But these two rotations are together equivalent to a single rota- 
tion from a to c, no matter what the relative positions of a, &, c may have 

* The terms minutes and seconds are derived from their Latin names, which 
are partes minutss primse and partes minutx secundss. At present there is 
a slight tendency among some authors to divide the degree decimally instead of 
into minutes and seconds. Still other authors use the degree and minute and 
divide the minutes decimally. Exercises involving both these systems will be 
found in the text. When the metric system was introduced at the end of the 
eighteenth century it was proposed to divide the right angle into 100 parts, called 
grades. The grade was divided into 100 minutes and the minute into 100 sec- 
onds. This system is used in some European countries, but not at all in America. 




146 MATHEMATICAL ANALYSIS [VI, § 101 

been. Hence, we have for any three half-lines a, 6, c issuing from a point 0, 
(1) (a&) + (6c) = (ac), (ab) + (&a) = 0, (ab) = (cb)-(ca). 

The proof of the last relation is left as an exercise. 
These relations are analogous to those of § 35 ; but an essential difference 
must be noted. Given two points A and 5 on a line, we may speak of the 
directed segment AB. The measure of AB is completely determined 
when A and B and the unit of measure are given. 
But if the half-lines a and b are given, the angle 
(ab) may be any angle generated by a rotation from 
a to 6. Such angles may be positive or negative and 
may involve, in addition to the minimum rotation 
from a to &, any number of complete revolutions. 
It is to be noted, however, that all possible determi- 
nations of the angle (ab) differ among themselves only by integral multi- 
ples of 360°. In other words, if 6 represents the smallest positive measure 
(in degrees) of an angle from a to b, then any determination of (ab) is 
given by the relation (ab) = 6 ± n- 360° (n an integer). The equality 
signs in relations (1) are then to be interpreted as meaning eqtial, except 
for multiples o/360°. 

If the position of the half-line h is determined by 
the angle di which it makes with a given horizontal line 
OX, and the position of another half-line h is deter- 
mined by the angle 62 which it makes with OX we have 

angle from Zi to ^2 = 60 — 61 , 
except for multiples of 360°. Why ? 

EXERCISES 

1. What angle does the minute hand of a clock describe in 2 hours and 
30 minutes ? in 4 hours and 20 minutes ? 

2. Suppose that the dial of a clock is transparent so that it may be read 
from both sides. Two persons stationed on opposite sides of the dial ob- 
serve the motion of the minute hand. In what respect will the angles de- 
scribed by the minute hand as seen by the two persons differ ? 

3. In what quadrants are the following angles : 87° ? 135°? - 326° ? 
540°? 1500°? -270°? 

4. In what quadrant is 6/2 if ^ is a positive angle less than 360° and in 
the second quadrant ? third quadrant ? fourth quadrant ? 

5. By means of a protractor construct 27°-}- 85° -f (—30°) -f 20°-|- (— 45°). 

6. By means of a protractor construct — 130° -\- 56° — 24°. 




VI, § 102] TRIGONOMETRIC FUNCTIONS 



147 



102. The Sine, Cosine, and Tangent of an Angle. We 

may now define three of the functions referred to in § 97. To 
this end let 9 = XOP (Fig. 80) be any directed angle, and let 



O X ^ 
Fig. 80 



w 



T^ 



US establish a system of rectangular coordinates in the plane 
of the angle such that the initial side OX of the angle is the 
positive half of the i»-axis, the vertex being at the origin and 
the i/-axis being in the usual position with respect to the 
aj-axis. Let the units on the two axes be equal. Finally, let 
P be any point other than on the terminal side of the angle 
6, and let its coordinates^ be («, y). The directed segment 
OP = r is called the distance of P and is always chosen posi- 
tive. The coordinates x and y are positive or negative accord- 
ing to the conventions previously adopted. We then define 
ordinate of P y 



The sine of 9 



The cosine of d = 



The tangent of Q = 



distance of P 

abscissa of P 
distance of P 

ordinate of P 



provided x =^ 0,* 



abscissa of P 

These functions are usually written in the abbreviated forms 
sin 0, cos 6, tan 6, respectively ; but they are read as " sine 0/^ 
" cosine 6/^ " tangent ^." It is very important to notice that 
the values of these functions are independent of the position 
of the point P on the terminal line. For let P{x', y') be any 
other point on this line. Then from the similar right triangles 
* Prove that x and y cannot be zero simultaneously. 



148 



MATHEMATICAL ANALYSIS 



[VI, § 102 



xyr* and a;'?/'r' it follows that the ratio of any two sides 
of the triangle xyr is equal in magnitude and sign to the 
ratio of the corresponding sides of the triangle xfy'r\ There- 
fore the values of the functions just defined depend merely 
on the angle 6. They are one-valued functions of 6 and are 
called trigonometric functions.^ 

Since the values of these functions are defined as the ratio 
of two directed segments, they are abstract numbers. They 
may be either positive, negative, or zero. Eemembering that r 
is always positive, we may readily verify that the signs of the 
three functions are given by the following table. 



Quadrant 

Sine 

Cosine 

Tangent 


l' 

+ 
+ 

+ 


2 

+ 


3 

+ 


4 



103. Values of the Functions for 45°, 135°, 225°, 315°. In 

each of these cases the triangle xyr is isosceles. Why? 
Since the trigonometric functions are independent of the 
position of the point P on the terminal line, we may choose 
the legs of the right triangle xyr to be of length unity, which 



^ j^l 



P' 



FiQ. 81 






gives the distance OP as V2. Figure 81 shows the four angles 



* Triangle xyz means the triangle whose sides are x,y,z. 

t Trigonometric etymologically means relating to the measurement 
of triangles. The connection of these functions with triangles will appear 
presently. 



VI, § 104] TRIGONOMETRIC FUNCTIONS 



149 



with, all lengths and directions marked. Therefore, 



sin 45'=-^, 

V2 



sin 135° 



sin 225^ 



V2' 



sin 315° = 



V2' 



vr 



cos 45°= ■" , 

V2' 


tan 45° = 1, 


cos 135°= ^ , 

V2 


tan 135° - - 1, 


cos 225°= ^ , 

V2 . 


tan 225° = 1, 


cos315°=-i:„ 


tan 315° = - 1. 



V2^ 



104. Values of the Functions for 30°, 150°, 210°, 330°. From 
geometry we know that if one angle of a right triangle con- 
tains 30°, then the hypotenuse is double the shorter leg, 
which is opposite the 30° angle. Hence if we choose the 
shorter leg (ordinate) as 1, the hypotenuse (distance) is 2, 



Jft' £n<^:^ 



v^ 



•VJ 






.^^' 



330 



Fig. 82 






and the other leg (abscissa) is V3. Figure 82 shows angles of 
30°, 150°, 210°, 330° with all lengths and directions marked. 
Hence we have 



sin 30° =i 




cos 30° = Y, 


tan 30°=-^, 
V3 


sin 150° =i 




cos 150° = -^, 


tan 150° = -—, 
V3 


sin 210° = - 


1 

2' 


cos 210°=-^, 


tan210°=-J:^, 
V3 


sin 330°=- 


1 

2' 


cos 330° = ^^, 


ton 330° = --^. 
V3 



150 



MATHEMATICAL ANALYSIS 



[VI, § 105 



105. Values of the Functions for 60°, 120°, 240°, 300°. It is 

left as an exercise to construct these angles and to prove that 

sin 60° = 
sin 120° = 
sin 240° = - 
sin 300° = - 

106. Applications. The angle which a line from the eye to 

an object makes with a horizontal line in the same vertical 
plane is called an angle of elevation or an angle of depression. 



V3 
2 ' 


cos 60° = L 

2 


tan 60°= a/3, 


V3 

> 

2 


cos 120° = -1, 


tan 120° -^ - V3, 


V3 
2 ' 


COS 240°=- ^, 


tan 240° = v'3. 


V3 

2 ' 


cos 300° =|, 
2 


tan300° = -V3. 





C^ 


Horizontal 


^.^^^"'""" 


Fia. 83 




Horizontal 


^^^ 



according as the object is above or below the eye of the observer 
(Fig. 83). Such angles occur in many examples. 

Example 1. A man wishing to know the distance between two points 
A and B on opposite sides of a pond, locates a point C on the land (Fig. 
84) such that ^C = 200 rd., angle G — Z^\ and angle B = 90°. Find the 
distance AB. 

^=sin(7. (Why?) 
AG 

AB=AC sin a 

= 200 . sin 30° 

= 200 . I = 100 rd. 

Example 2. Two men stationed at points A and C 800 yd. apart and 
in the same vertical plane with a balloon B, observe simultaneously the 
angle of elevation of the balloon to be 30° and 45° respectively. Find the 
height of the balloon. 




Solution : 



Fia. 84 



VI, § 106] TRIGONOMETRIC FUNCTIONS 



151 



Solution : Denote the height of the balloon DB by y, and let DC = x 
then AD = 800 - x. 




tan 45° = 1, we have 1 = ^ 

X 



Since 
and since tan 30° = 1/ V3, we have 



V3 800 - X 
Therefore oc = y and 800 — x = y \' 8. 

800 



Solving these equations for y, we have 



V3+ 1 



292.8 yd. 



EXERCISES 

1. In what quadrants is the sine positive ? cosine negative ? tangent 
positive ? cosine positive ? tangent negative ? sine negative ? 

2. In what quadrant does an angle lie if 

(a) its sine is positive and its cosine is negative ? 
(6) its tangent is negative and its cosine is 'positive ? 

(c) its sine is negative and its cosine is positive ? 

(d) its cosine is positive and its tangent is positive ? 

3. Which of the following is the greater and why : sin 49° or cos 49° ? 
sin 35° or cos 35° ? 

4. If 6 is situated between 0° and 360°, how many degrees are there in 
6 if tan ^ = 1 ? Answer the similar question for sin = ^ ; tan ^ = - 1. • 

5. Does sin 60° = 2 • sin 30° ? Does tan 60° = 2 • tan 30° ? AYhat 
can you say about the truth of the equality sin 2 ^ = 2 sin ^ ? 

6. The Washington Monument is 555 ft. high. At a certain place in 
the plane of its base, the angle of elevation of the top is 60°. How far is 
that place from the foot and from the top of the tower ? 

7. A boy \vhose eyes are 5 ft. from the ground stands 200 ft. from a 
flagstaff. From his eyes, the angle of elevation of the top is 30°. How 
high is the flagstaff ? 



152 MATHEMATICAL ANALYSIS [VI, § 106 

8. A tree 38 ft. high casts a shadow 88 ft. long. What is the angle 
of elevation of the top of the tree as seen from the end of the shadow ? 
How far is it from the end of the shadow to the top of the tree ? 

9. From the top of a tower 100 ft. high, the angle of depression of 
two stones, which are in a direction due east and in the plane of the base, 
are 45° and 30° respectively. How far apart are the stones ? 

^ns. 100(V3-l)=73.2ft. 

10. Find the area of the isosceles triangle in which the equal sides 
10 inches in length Include an angle of 120°. Ans. 25 VS = 43.3 sq. in. 

11. Is the formula sin2 ^ = 2 sin ^cos ^ true when = 30°? 60°? 
120^^^^ ? 

12. From a figure prove that sin 117° = cos 27°. 

13. Find the tangent of the angle which the line joining the points 
(^1) yi)i and (X2, Vi) makes with the aj-axis, assuming the units on the 
two axes to be equal. Compare your answer with the definition of slope 
in §§ 50 and 53. 

14. Determine whether each of the following formulas is true when 
e = 30°, 60°, 150°, 210° : 

1 + tan2 d 



1 + 



cos'^ d 

1 1 



tan2 d sin2 0' 
sin2 + cos2 = 1. 

16. Let Pi(a:i, yi) and P2(a;2, 2/2) be any two points the distance be- 
tween which is r (the units on the axes being equal). If is the angle 
that the line P1P2 makes with the cc-axis, prove that 

X2-X1 J y2 - yi ^ 2 r. 

cos 6 sin 6 

107. Computation of the Value of One Trigonometric 
Function from that of Another. 

Example 1. Given that sin = f , find the 
6y\ hs5 values of the other functions. 

XB I _J__1^^ Since sin 6 is positive, it follows that d is 

an angle in the first or in the second quad- 
rant. Moreover, since the value of the sine 
is I, then y = Z - k and r = 6 •% where k is 
any positive constant different from zero. (Why ?) It is, of course, 
immaterial what positive value we assign to A;, so we shall assign the 



Fia. 86 



VI, § 108] TRIGONOMETRIC FUNCTIONS 



153 



value 1. We know, however, that the abscissa, ordinate, and distance 
are connected by the relation x"^ + y^ = r^, and hence it follows that 
X = ± 4. Fig. 86 is then self-explanatory. Hence we have, for the first 
quadrant, sin ^=f, cos^=|, and tan^=|; for the second quadrant, 
sin ^ = |, cos ^ = — I, tan 5 = — |. 



Example 2. Given that sin d = f^ and that tan d 
is negative, find the other trigonometric functions of 
the angle 6. 

Since sin 6 is positive and tan 6 is negative, 6 must 
be in the second quadrant. We can, therefore, con- 
struct the angle (Fig. 87), and we obtain sin ^ = j\, 



:^1_ 



Fig. 87 



108. Computation for Any Angle. Tables. The values of 
the trigonometric functions of any angle may be computed by 

the graphic method. For 
example, let us find the 
trigonometric functions of 
35°. We first construct 
on square ruled paper, 
by means of a protractor, 
an angle of 35° and choose 
a point P on the ter- 
minal line so that OF 
shall equal 100 units. 
Then from the figure we 
find that 0M= 82 units 
iind MP = 57 units. 
Fig. 88 Therefore 

sin 35° = ^ = 0.57, cos 35° = -^\\ = 0.82, tan 35° = || = 0.70. 
The tangent may be found more readily if we start by tak- 
ing OA — 100 units and then measure AB. In this case, 
AB = 70 units and hence tan 35° = t% = ^•'^^• 

It is at once evident that the graphic method, although 



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::::! 


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!■■■■■ sSsssssss 




:::: 


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;| 






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iiillll 




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mt 


k 


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iii? 


lliiii 


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+ : 


+£ 


u:i:i::i:i:K:i 




:±i: 






i 




ilss 


:: 


:; 




:l::Kypi::::: 




-1 


1 


g 


■4- 
it 


i 


ii 




1 


1 


== 


II 




llil 


::: 




■:i 


1 


s 




i| 




i: 


Iwif 


S== 




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illf 




±: 


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10 no 30 40 



80 90 too 



154 



MATHEMATICAL ANALYSIS [VI, § 108 




10 

Fig. 



20 30 iO SO CO 70 60 90 100 

89. — Graphical Table of Trigonometric Functions 



VI, § 108] TRIGONOMETRIC FUNCTIONS 155 

simple, gives only an approximate result. However, the values 
of these functions have been computed accurately by methods 
beyond the scope of this book. The results have been put in 
tabular form and are known as tables of natural trigonometric 
functions. These tables with an explanation of their use will 
be found in any good set of mathematical tables.* In order 
to solve several of the following exercises it is necessary to 
make use of such tables. 

Figure 89 makes it possible to read off the sine, cosine, or 
tangent of any angle between 0° and 90° with a fair degree of 
accuracy. The figure is self-explanatory. Its use is illustrated 
in some of the following exercises. 

EXERCISES 

Find the other trigonometric functions of the angle 6 when 

1. tan^ = -3. 3. cos ^= if. 6. sin^ = f. 

2. sin^=-|. 4. tan^=4. 6. cos^=-^ 

7. sin ^ = I and cos 6 is negative. ■ _, 

8. tan 6 = 2 and sin 6 is negative. 

9. sin e = — ;^ and tan 6 is positive. 

10. cos ^ = f and tan 6 is negative. 

11. Can 0.6 and 0.8 be the sine and cosine, respectively, of one and 
the same angle ? Can 0.5 and 0.9 ? Ans. Yes ; no. 

12. Is there an angle whose sine is 2 ? Explain. 

13. Determine graphically the functions of 20°, 38°, 70°, 110°. Check 
your results by the tables of natural functions. 

U. From Fig, 89, find values of the following : 

sin 10°, cos 50°, tan 40°, sin 80°, tan 70°, cos 32°, tan 14°, sin 14°. 

15. A tower stands on the shore of a river 200 ft. wide. The angle of 
elevation of the top of the tower from the point on the other shore exactly 
opposite to the tower is such that its sine is |. Find the height of the 
tower. 

* See, for example, The Macmillan Tables, which will be referred to 
in connection with this book. 



156 MATHEMATICAL ANALYSIS [VI, § 108 

16. From a ship's masthead 160 feet above the water the angle of de- 
pression of a boat is such that the tangent of this angle* is ^j. Find tlie 
distance from the boat to the ship. Ans. 640 yards. 

17. A certain railroad rises 6 inches for every 10 feet of track. What 
angle does the track make with the horizontal ? 

18. On opposite shores of a lake are two flagstaffs A and B. Per- 
pendicular to the line AB and along one shore, a line BC = 1200 ft. is 
measured. The angle ACB is observed to be 40° 20'. Find the distance 
between the two flagstaffs. 

19. The angle of ascent of a road is 8°. If a man walks a mile up the 
road, how many feet has he risen ? 

20. How far from the foot of a tower 150 feet high must an observer, 
6 ft. high, stand so that the angle of elevation of its top may be 23°. 5 ? 

21. From the top of a tower the angle of depression of a stone in the 
plane of the base is 40° 20'. What is the angle of depression of the stone 
from a point halfway down the tower ? 

22. The altitude of an isosceles triangle is 24 feet and each of the equal 
angles contain 40° 20'. Find the lengths of the sides and area of the 
triangle. 

23. A flagstaff 21 feet high stands on the top of a cliff. From a point 
on the level with the base of the cliff, the angles of elevation of the top 
and bottom of the flagstaff are observed. Denoting these angles by a 
and /3 respectively, find the height of the cliff in case sin a ■=. ^ and 
cos/3 = H. Ans. 76 feet. 

24. A man wishes to find the height of a tower CB which stands on a 
horizontal plane. From a point A on this plane he finds the angle of ele- 
vation of the top to be such that sin CAB = f . From a point A' which 
is on the line AC and 100 feet nearer the tower, he finds the angle of 
elevation of the top to be such that tan CA'B — f . Find the height of the 

. tower. 

25. Find the radius of the inscribed and circumscribed circle of a regu- 
lar pentagon whose side is 14 feet. 

26. If a chord of a circle is two thirds of the radius, how large an 
angle at the center does the chord subtend ? 

27. A boy standing a feet behind and opposite the middle of a football 
goal observes the angle of elevation of the nearer crossbar to be a, and 
the angle of elevation of the farther crossbar to be p. Prove that the 
length of the field is a [tan a — tan /3]/tan /3. 




VI, § 109] TRIGONOMETRIC FUNCTIONS 157 

109. The Sine Function. Let us trace in a general way the 
variation of the function sin ^ as ^ increases from 0° to 360°. 
For this purpose it will be convenient to think of the distance 
r as constant, from which it follows that 
the locus of P is a circle. When = 0°, the 
point P lies on the x-axis and hence the 
ordinate is 0, i.e. sin 0° = 0/r = 0. As ^ 
increases to 90°, the ordinate increases 
until 90° is reached, when it becomes equal 
to r. Therefore, sin 90° = r/r = 1. As ^ p^^ ^ 

increases from 90° to 180°, the ordinate de- 
creases until 180° is reached, when it becomes 0. Therefore 
sin 180° = 0/r = 0. . As ^ increases from 180° to 270°, the ordi- 
nate of P continually decreases algebraically and reaches its 
smallest algebraic value when = 270°. In this position the 
ordinate is — r and sin 270° = — r/r = — 1. When enters 
the fourth quadrant, the ordinate of P increases (algebraically) 
until the angle reaches 360°, when the ordinate becomes 0. 
Hence, sin 360° = 0. It then appears that : 

as 6 increases from 0° to 90°, sin increases from to 1 ; 

as 6 increases from 90° to 180°, sin decreases from 1 to ; 

as 6 increases from 180° to 270°, sin 6 decreases from to — 1 ; 

as 6 increases from 270° to 360°, sin $ increases from — 1 to 0. 
It is evident that the function sin 6 repeats its values in the 
same order no matter how many times the point P moves 
around the circle. We express this fact by saying that the 
function sin 6 is periodic and has a period of 360°. In symbols 
this is expressed by the equation 

sin [6 + n . 360°] = sin 9, 

where n is any positive or negative integer. 

The variation of the function sin 6 is well shown by its 



158 



MATHEMATICAL ANALYSIS 



[VI, § 109 



graph. To construct this graph proceed as follows : Take a 
system of rectangular axes and construct a circle of unit radius 




Fig. 91 
with its center on the aj-axis (Fig. 91). Let angle XOP = 0. 
Then the values of sin 6 for certain values of 6 are shown in 
the unit circle as the ordinates of the end of the radius drawn 
at an angle 0. 



d 





30° 


45° 


60° 


90° 


... 


sin^ 





M,Pi 


M2P2 


MsPs 


M,P, 





Now let the number of degrees in ^ be represented by dis- 
tances measured along OX. At a distance that represents 30° 
erect a perpendicular equal in length to sin 30° ; at a distance 
that represents 60° erect one equal in length to sin 60°, etc. 
Through the points 0, Pi, P^,-" draw a smooth curve'; this 
curve is the graph of the function sin 6. 

If from any point P on this graph a perpendicular PQ is 
drawn to the a;-axis, then QP represents the sine of the angle 
represented by the segment OQ. 

Since the function is periodic, the complete graph extends 
indefinitely in both directions from the origin (Fig. 92). 

Y 




VI, § 110] TRIGONOMETRIC FUNCTIONS 



159 



110. The Cosine Function. By arguments similar to those 
used in the case of the sine function we may show that : 
as 6 increases from 0°to 90°, the cos decreases from 1 to ; 
as $ increases from 90° to 180°, the cos 9 decreases from to — 1 ; 
as increases from 180° to 270°, the cos 6 increases from — 1 to ; 
as increases from 270° to 360°, the cos increases from to 1. 

The graph of the" function is readily constructed by a method 




Fig. 93 



similar to that used in case of the sine function. This is 
illustrated in Fig. 93. 

The complete graph of the cosine function, like that of the 
sine function, will extend indefinitely from the origin in both 



7 




^^ /^ 


\ AA 


Vy '' 


\ / \-^ 


-i 


^=cosx 



Fig. 94 



directions (Fig. 94). Moreover cos 0, like sin 0, is periodic and 
has a period of 360°, i.e. 

cos [6 + n . 360°] = cos e, 

where n is any positive or negative integer. 



160 



MATHEMATICAL ANALYSIS 



[VI, § 111 




Fig. 95 



111. The Tangent Function. In order to trace the varia- 
tion of the tangent function, consider a circle of unit radius 
with its center at the origin of a system of rectangular axes 
(Fig. 95). Then construct the tangent to 
this circle at the point M{1, 0) and let P 
denote any point on this tangent line. If 
angle MOP = 0, we have tan 6 = MP/OM 
= MP /I — MP, i.e. the line MP represents 
tana 

Now when 6 = 0°, MP is 0, i.e. tan 0° is 0. 

As the angle 6 increases, tan 6 increases. As 

6 ap]3roaches 90° as a limit, MP becomes 

infinite, i.e. tan 6 becomes larger than any number whatever. 

At 90° the tangent is undefined. It is sometimes convenient 

to express this fact by writing 

tan 90° = 00. 

However we must remember that this is not a definition for 
tan 90°, for oo is not a number. This is merely a short way of 
saying that as $ approaches 90°, tan becomes infinite and 
that at 90° tan 6 is undefined. See § 36. 
Thus far we have assumed 6 to be an 
acute angle approaching 90° as a limit. 
Now let us start with 6 as an obtuse angle 
and let it decrease towards 90° as a limit. 
In Fig. 96 the line MP' (which is here 
negative in direction) represents tan 9. 
Arguing precisely as we did before, it is 
seen that as the angle 6 approaches 90° 
as a limit, tan 6 again increases in magnitude beyond all 
bounds, i.e. becomes infinite, remaining, however, always 
negative. 




Fig. 96 



VI, § 111] 



TRIGONOMETRIC FUNCTIONS 



161 



We then have the following results. , 

(1) When 6 is acute and increases toward' 90° as a limit, 
tan always remains positive but becomes infinite. At 90° 
tan is undefined. 

(2) When 6 is obtuse and decreases towards 90° as a limit, 
tan 6 always remains negative but becomes infinite. At 90° 
tan 6 is undefined. 

It is left as an exercise to finish tracing the variation of the 
tangent function as 6 varies from 90° to 360°. Note that 
tan 270°, like tan 90°, is undefined. In fact tan n • 90° is unde- 
fined, if n is any odd integer. 




360° X 



Fia. 97 



To construct the graph of the function tan 9 we proceed 
along lines similar to those used in constructing the graph of 
sin e and cos 0. The following table together with Fig. 97 
illustrates the method. 



d 


0° 


30° 


45° 


60° 


90° 


120° 


136° 


150° 


180° 


210° 


tan^ 





MPi 


MP2 


MPz 


undefined 


MPi 


MP^ 


MP^ 


3/P7=0 


MPi 



162 MATHEMATICAL ANALYSIS [VI, § 111 

It is important to notice that tan 6y like sin 9 and cos 6j is 
periodic, but its period is 180°. That is 

tan (6 + 1 • i8o°)= tan 6, 

where n is any positive or negative integer. 

EXERCISES 

1. What is meant by the period of a trigonometric function ? 

2. What is the period of sin 6 ? cos 6 ? tan d ? 

3. Is sin d defined for all angles ? cos d ? 

4. Explain why tan 6 is undefined for certain angles. Name four 
angles for which it is undefined. Are there any others ? 

5. Is sin (^ + 360^) = sin d ? 

6. Is sin {e + 180°) =sin^? 

7. Is tan {d + 180°) = tan ^ ? 

8. Is tan {6 + 360°) = tan ? 

Draw the graphs of the following functions and explain how from the 
graph you can tell the period of the function : 

9. sin^. 11. tan^. 13. -^• 

cos d 

10. cosd. 12. — — 14. ^ . 

sin d tan 6 

Verify the following statements : 

16. sin 90° + sin 270° = 0. 18. cos 180° + sin 180° = - 1. 

16. cos 90° + sin 0° = 0. 19. tan 360° + cos 360° = 1. 

17. tan 180° + cos 180°=- 1. 20. cos90°-f tan 180°-sin270° = l. 

21. Draw the graphs of the functions sin ^, cos 6, tan ^, making use of 
a table of natural functions. See p. 638. 

22. Draw the curves y = 2 sin ^ ; y = 2 cos ; y = 2 tan 6. 

23. Draw the curve j/ = sin + cos 6. 

24. From the graphs determine values of d for which sin ^ = ^ ; sin 6 
= 1 ; tan <? = 1 ; cos = ^ ; cos ^ = 1. 



VI, § 112] TRIGONOMETRIC FUNCTIONS 



163 




Fig. 98 



112. Polar Coordinates. It is convenient at this point to 
introduce a new way of locating the position of a point in a 
plane, and of representing the graph of a function. To this end 
(Fig, 98) let OA be a directed line in the plane which we shall 
call the initial line or the polar axis. 
This line is usually drawn horizontally 
and directed to the right. The point 
is called the pole or the origin. Let P 
be any point in the plane and draw the 
line OP. The position of P is then 
located completely if we know the angle ^OP=^and the dis- 
tance OP=:p. The two numbers (p, 9), called respectively the 
radius vector and the vectorial angle, are known as the polar 
coordinates of the point P. 

In Fig. 98 we have represented a case in which and f> are 
both positive. Either ot p or both may be negative under 
the following conventions. The angle is positive or negative 
according to the direction of its rotation, as in § 98. The 
positive direction on OP is the direction from along the 
terminal side of the angle 0, i.e., it is the direction into which 
OA is rotated by a rotation through the angle 0. 
With these conventions a point P whose polar coordinates 
(/), 0) are given is completely de- 
termined. Figure 99 shows points 
whose polar coordinates are (2, 30°), 
(-2, 30°), (2, -30°), and (-2, 
— 30°). It will be noted that, if p is 
positive, P is on the terminal side of 
$, while if p is negative, P is on the 
terminal side produced through 0. 
On the other hand, a given point P has an unlimited number of 
polar coordinates (p, 6). Even if we confine ourselves to angles 



i-s,-so') 



{2,30') 




(-s,so ) 



i2y-30°) 



Fig. 99 



164 MATHEMATICAL ANALYSIS [VI, § 112 

in absolute value less than 360°, a point lias in general /owr dif- 
ferent sets of polar coordinates. Fig. 100 shows that the same 



(e.so'). 




point P may be designated by any one of the pairs of values 
(2, 30°), (2, - 330°), (- 2, 210°), and (- 2, - 150°). 

EXERCISES 

1. Locate the points whose polar coordinates have the following values : 
(4, 30°), (-2, 45<^), (-3, -60°), (2, -160°), (3, -90°), (2, 180°), 
(-2, 0°), (0, 90°), (-2, 180°), (- 3, 270°). 

2. For each of the points in Ex. 1, give all other sets of polar coordi- 
nates for which 6 is in absolute value less than 360°. 

3. What exceptions are there to the statement " 6 being confined to 
angles in absolute value less than 360°, every point has four and only 
four distinct sets of polar coordinates " ? 

4. Where are all the points for which ^ is a given constant ? 

5. Where are all the points for which p is a given constant ? 

113. Graphs in Polar Coordinates. Polar coordinates may 
be used to represent the graph of a given function, in a way 
quite similar to that in the case of rectangular coordinates. 
Fig. 101 gives an example in which the idea of polar coor- 
dinates is used in practice. In this example the ^-scale rep- 
resents time, the p-scale represents temper ature.*^ Some forms 
of self-recording hygrometers employ the same idea. 

* It will be noted that in this example the radius vector is measured along 
a circular arc instead of along a straight line. This is due to the mechanical 
(construction of the instrument. Cf. footnote, p. 9. The fundamental idea is, 
nevertheless, that of polar coordinates. 



VI, § 1131 TRIGONOMETRIC FUNCTIONS 



165 ' 



In plotting the graph of a function in polar coordinates we 
proceed as in the case of rectangular coordinates. A table of 




Fig. 101 



corresponding values of the variable 6 and the function p is 



166 



MATHEMATICAL ANALYSIS 



[VI, § 113 



constructed. Each such pair of values is then plotted as a 
point, and a curve drawn through these points. 

Example. Plot in polar coordinates the graph of p = sin 6. We ob- 
tain the table below. Figure 102 exhibits the corresponding points, with 



(1.90) 



e 


p = sin ^ 


0° 


.00 


30^ 


.50 


46° 


.71 


60° 


.87 


90° 


1.00 


120° 


.87 


135° 


.71 


150° 


.50 


180° 


.00 


210° 


- .50 


225° 


- .71 


240° 


- .87 


270° 


- 1.00 


300° 


- .87 


315° 


- .71 


330° 


- .50 


360° 


.00 




a curve drawn through them. Observe that each point serves to represent 
two pairs of corresponding values. Thus the pairs (^, 30°) and ( ~ i, 210°) 
are represented by the same point. This curve suggests a circle, of diame- 
ter unity, tangent to the polar axis at the origin. 

114. The Graph of sin 6 and cos 6 in Polar Coordinates. 

We may now prove : 

The graph, in polar coordinates, of the function p — sin $ is a 
circle of diameter unity, tangent to the polar axis at the origin. 

Let P (p, 6) be any point on such a circle (Fig. 103). Then, 
for any value 6 in the first quadrant 



OA 1 



or 



p = sin ^. 



VI, § 114] TRIGONOMETRIC FUNCTIONS 



167 



Conversely, if p = sin 6, the point P is on the circle. Why ? 
A similar proof, which is left as an exercise, may be given 
when 6 is in the second, third, or fourth quadrants (Fig, 104). 
Similarly, we may prove : 





FiQ. 1(M 



Hie graph of 



p = GO8 



in polar coordinates is a circle of diameter unity, passing through 
the pole and having its center on the polar axis. 

The proof of this statement is left as an exercise. See Figs. 
105, 106. 

On account of their simplicity, the polar graphs of sin $ and 
cos 6 are very serviceable. It is for this reason that we have 





FiQ. 105 



Fig. 106 



introduced them at this point. Polar coordinates will be dis- 
cussed again, particularly in Chapter XIV, and incidentally 
in other chapters. 



168 MATHEMATICAL ANALYSIS [VI, § 114 

EXERCISES 

1. From Fig. 101, find the temperature at 9 p.m. on Tuesday ; at 3 p.m. 
on Monday. When was the temperature a maximum ? a minimum ? 

2. Plot in polar coordinates the graph representing the variation in 
temperature given in Ex. 1, p. 16. 

3. Plot the graph in polar coordinates of the function p = tan d. Why 
is this graph not convenient to represent the function tan 6 ? 

4. Prove that the graph, in polar coordinates, of /> = a cos ^ is a circle 
of diameter a, passing through the origin and w^ith its center on the polar 
axis. 

5. Prove a theorem regarding the graph of p = a sin d, 

115. Other Trigonometric Functions. The reciprocals of 
the sine, the cosine, and the tangent of any angle are called, 
respectively, the cosecant, the secant, and the cotangent of 
that angle. Thus, 

cosecant = distance of P ^ r ( i^^^ ^ o). 
ordinate of P y ^^ ^ ^ 

, /J distance of P r , ., , ^^.^ 
secant 6 = — — : = - (provided x^O), 

abscissa of P x 

. , /, abscissa of P a? , • i •, , r^. 

cotangent 6 = — = - (provided y^O). 

ordinate oi P y 

These functions are written esc 6, sec 0, ctn d. From the 
definitions follow directly the relations 

csce = ^i-, sece = -, ctn0 = 



sin 6' COS0' tan 9' 

or 

esc ^ • sin ^ = 1, sec ^ • cos ^ = 1, ctn • tan ^ = 1. 

To the above functions may be added versed sine (written versin), 
the coversed sine (written coversin), and the external secant (written 



VI, § 116] TRIGONOMETRIC FUNCTIONS 



169 



exsec), which are defined by the equations versin 6 = 1 — coa 6, coversin e 
= 1 — sin 6, and exsec 6 = sec d — 1. 



It is left as an exercise to trace the variation of esc By sec 6, 
ctn 0, as varies from 0° to 360°. Be careful to note tliat 
ctn 0°, ctn 180°, esc 0°, esc 180°, sec 90°, sec 270° are undefined. 
Why? 

116. The Representation of the Functions by Lines. We 

have seen in §§ 109-111, that if we take a unit circle we may 
represent sin 9, cos 0, and tan by means of lines. We will 
now extend this representation to include esc 6, sec 6, ctn 6. 




Fig. 107 



Figure 107 shows the functions in a unit circle for an angle 
6 in the first quadrant. We have 



MF = sin e 
OM=Gos6 



AT 
BS 



tan^ 
ctn^ 



Or=sec^ 
OS = CSC $ 



Draw similar figures for angles in each of the other quad- 
rants. The points may be so labeled that the results given 
for the first quadrant hold in any quadrant. 



170 MATHEMATICAL ANALYSIS [VI, § 117 

117. Relations among the Trigonometric Functions. As 

one might imagine, the six trigonometric functions sine, cosine, 
tangent, cosecant, secant, cotangent are connected by certain 
relations. We shall now find some of these relations. 
From Fig. 80 (§ 102) it is seen that for all cases we have 

(1) • 0:2 4- 2/2 = ,.2_ 

If we divide both sides of (1) by r^, we have 

^ -f- ^ = 1 (by hypothesis r^0)\ 

or 

sin2 + cos'e = l. 

Dividing both sides by a;^, we have 

1 + ^ = ^ iiix^O), 
x^ x^ 

Therefore 

1 + tan^ 6 = sec2 6. 



Similarly dividing both sides of (1) by if gives 



^ + 1 = S (if.'/^O); 



or 

ctn2 e + 1 = csc2 e. 

Moreover, we have 

X X COS 
r 

and, similarly, 

ctne=5?i|. 
sm6 



VI, § 118] TRIGONOMETRIC FUNCTIONS 171 

118. Identities. By means of the relations just proved 
any expression containing trigonometric functions may be 
put into a number of different forms. It is often of the 
greatest importance to notice that two expressions, although 
of a different form, are nevertheless identical in value. (See 
§ 47 for the definition of an identity.) 

The truth of an identity is usually established by reducing 
both sides, either to the same expression,- or to two expres- 
sions which we know to be identical. The following examples 
will illustrate the methods used. 

Example 1. Prove the relation sec^ d + csc^ d = sec^ d csc^ d. 
We may write the given equation in the form 

sec2 d csc2 5, 



or 



which reduces to 



cos'^ d sin^ 



?Hl!i±^2sif = sec2ecsc2^, 
cos2 dsin^d 



sec2 d csc2 d, 



cos2 e sin2 d 



sec2 d csc2 d = sec2 d csc^ 6. 



Since this is an identity, it follows, by retracing the steps, that the 
given equality is identically true. 

Both members of the given equality are undefined for the angles 0°, 90°, 

180°, 270°, 360° or any multiples of these angles. 

cos*^ 6 

Example 2. Prove the identity 1 + sin ^ = ■, — -• 

1 — sm ^ 

Since cos^ ^ = 1 — sin2 ^, we may write the given equation in the form 

1 + sin = ^ ~ ^^"^ ^ or 1 + sin d = 1 + sin 6. 
1 - sin ^ 

As in Example 1, this shows that the given equality is identically true. 

The right-hand member has no meaning when sin = 1, while the left- 
hand member is defined for all angles. We have, therefore, proved that 
the two members are equal except for the angle 90° or (4 ji -|- 1) 90°, where 
n is any integer. 



172 MATHEMATICAL ANALYSIS [VI, § 118 

The formulas of § 117 may be used to solve examples of the 
type given in § 107. 

Example 3. Given that sin d = /^ and that tan 6 is negative, find the 
values of the other trigonometric functions. 

Since sin^ d + cos^ ^ = 1, it follows that cos ^ = ± ^|, but since tan 6 is 
negative, 6 lies in the second quadrant and cos 6 must be — ||. More- 
over, tlie relation tan d = sin 0/ cos 6 gives tan 6 =— j%. The reciprocals 
of these functions give sec ^ = — |f, esc 6 = ^^-^ ctn ^ = — -^. 

EXERCISES 

1. Define secant of an angle ; cosecant; cotangent. 

2. Are there any angles for which the secant is undefined ? If so, 
what are the angles ? Answer the same questions for cosecant and co- 
tangent. 

3. Define versed sine ; coversed sine. 

4. Complete the following formulas : 

sin20 + cos2 0=? l + tan2^ = ? l+ctn2^=? tan5=? 
Do these formulas hold for all angles ? 

5. In what quadrants is the secant positive ? negative ? the cosecant 
positive ? negative ? cotangent positive ? negative ? 

6. Is there an angle whose tangent is positive and whose cotangent is 
negative ? 

7. In what quadrant is an angle situated if we know that 

(a) its sine is positive and its cotangent is negative ? 
(6) its tangent is negative and its secant is positive ? 
(c) its cotangent is positive and its cosecant is negative ? 

8. Express sin^ ^ + cos ^ so that it shall contain no trigonometric 
function except cos 6. 

9. Transform (1 + ctn^ 6) esc so that it shall contain only sin 6. 

10. Which of the trigonometric functions are never less than one in 
absolute value ? 

11. For what angles is the following equation true : tan = ctn ? 

12. How many degrees are there in when ctn ^ = 1 ? ctn^ = — 1 ? 
sec ^ = \/2 ? CSC = \/2 ? 

13. Determine from a figure the values of the secant, cosecants **»^ 
cotangent of 30°, 160^, 2W\ 330°. 



VI, § 118] TRIGONOMETRIC FUNCTIONS 



173 



14. Determine from a figure the values of the secant, cosecant, and 
cotangent of 45% 135°, 226% 316°. 

16. Determine from a figure the values of the sine, cosine, tangent, 
secant, cosecant, and cotangent of 60^, 120°, 240°, 300°. 

16. Show that the graphs of the function sec e, esc ^, ctn d have the 
forms indicated in the adjacent figures. 




Prove the following identities and state for each the exceptional values 
of the variables, if any, for which one or both members are undefined : 

17. cos d tan d = sin d. 

18. sin 6 ctn Q = cos 6. 
1 + sin g _ cos5 

cos 6 



19 



1 — sin g 

20. sin2 6 — cos2 6 = 2 sin2 5-1. 

21. ( 1 - sin2 5) csc2 6 = ctn2 ^. 

22. tan 6 + ctn 5 = sec 5 esc 0. 

23. [x sine-t y cos ey -h [xcosd-y sin 6^ = x^ + ^. 

CSC 6 



24. 



= cos 0. 



tan + ctn 
26. 1 - ctn* = 2 csc2 - esc* 0. 

26. tan2 5-sin2 5 = tan2 5sin2g. 

27. 2(1 + sin 0) ( 1 + cos 0) = (1 + sin + cos ey. 

28. sin» + cos« 5 = 1—3 sin2 cos2 6. 

CSC 5 



29. _^!5^ + 



30. 



CSC 5 — 1 CSC 5 + 1 

1 — tan 9 _ Ctn g - 1 
1 + tan 5 ctD 5 -f 1 



= 2 sec2 e. 



174 MATHEMATICAL ANALYSIS [VI, § 118 

31. [1 + tan e + sec e][l + ctn e — esc ^] = 2. 

32. (taiid + sec^)2 = LiLSEi. 
^ ^ 1 - sin ^ 

33. CSC* ^ (1 - cos* d)-2 ctn2 ^ = 1. 

34. (tan d — ctn ^)sin ^ cos = 1 - 2 cos^ e. 

36. sec g - tan j ^ i _ 2 sec g tan ^ + 2 tan^ d. 
sec + tan 6 

36. ^-^?i«±J^^5j = tanatan/3. 
ctn a + ctn /3 

37. sin (sec d + esc ^) — cos (sec ^ — esc 6) = sec ^ esc d. 

Find algebraically the other trigonometric functions of the angle 
when 

38. ctn = 4: and sin is negative. 

39. sin = I and sec is positive. 

40. sec ^ = 2 and tan is negative. 

41. CSC ^ = — 5 and ctn is positive. 

119. Trigonometric Equations. An identity, as we have 
seen (§ 47), is an equality between tvro expressions which is 
satisfied for all values of the variables for which both expres- 
sions are defined. If the equality is not satisfied for all 
values of the variables for which each side is defined, it is 
called a conditional equality, or simply an equation. Thus 
1 — cos ^ = is true only if ^ = w • 360°, where n is an integer. 
To solve a trigonometric equation, i.e. to find the values of 
for which the equality is true, we usually proceed as follows. 

1. Express all the trigonometric functions involved in terms 
of one trigonometric function of the same angle. 

2. Find the value (or values) of this function by ordinary 
algebraic methods. 

3. Find the angles between 0° and 360° which correspond to 
the values found. These angles are called particular solutions. 

4. Give the general solution by adding n • 360°, where n is 
any integer, to the particular solutions. 



VI, § 119] TRIGONOMETRIC FUNCTIONS 175 

Example 1. Find d when sin 5 = ^. 

The particular solutions are 30° and 150°. The general solutions are 
30° + w . 360°, 150° + n • 360°. 

Example 2. Solve the equation tan ^ sin ^ — sin ^ = 0. 

Factoring the expression, we have sin 6 (tan^— 1) = 0. Hence we 
have sin ^ = 0, or tan ^—1=0. Why ? 

The particular solutions are therefore 0°, 180°, 45°, 225°. The general 
solutions are n • 360°, 180° + w • 360°, 45° + n . 360°, 225° + n . 360°. 

Example 3. Find d when tan 6 + ctn 6 = 2. 
The given equation may be written 

tan (? + —^ = 2, 

tan d 

or 

tan2 ^-2 tan ^+1 = 0; 
therefore 

(tan ^ - 1)2 = 0, or tan tf = 1. 

It follows that 6 = 46° or 225° ; or, in general, 

^ = 46° + n . 360° or 226° + n • 360°. 

EXERCISES 
Give the particular and the general solutions of the following equations : 

1. sin d = ^. 9. tan ^ = - 1. 

2. sin 5 = - ^. 10. ctn ^ = - 1. 

3. cos e= ^. 11. tan 6=1. 

4. cos 5 = — ^. 12. ctn ^ = 1. 
6. sec 6 = 2. 13. tan2 6 = 3. 

6. sec ^ = — 2. 14. sin ^ = 0. 

7. CSC 6 = 2. 15. cos ^ = 0. 

8. esc ^ = - 2. 16. tan 6 = 0. 

Solve the following equations giving the particular and the general 
solutions in each case : 

17. sin d = cos 6. Ans. 45°, 226° ; 45° + n • 360°, 226° + n • 360°. 

18. tan^^H- 2sec2^ = 6. 

19. 6 sin ^ + 2 cos2 6 = 5. Ans. 90° ; 90° + n • 360°. 

20. cos2 ^ + 5 sin = 7. 



176 



MATHEMATICAL ANALYSIS [VI, § 119 



21. 4 sin 5 — 3 esc ^ = 0. 

22. 2 sin 6 cos^ 5 = sin 5. 

23. cos 5 + sec 5 = f . 

24. 2 sin = tan d. Ans. Particular solutions : 0'', 180°, 60°, 300°. 

25. 3 sin ^ + 2 cos ^ = 2. 

26. 2cos2 ^-1 = 1- sin2 e . 

120. The Trigonometric Functions of — 9. Draw the angles 
and — 9, where OP is the terminal line of and OP' is the 
terminal line of — 0. Figure 108 shows an angle in each of 



P P 

X V 

p7> 




^ ^ 



Fig. 108 



X 



V X 



the four quadrants. We shall choose OP — OP' and {x, y) as 
the coordinates of P and (x\ y') as the coordinates of P'. In 
all four figures 



Hence 



x' = ic, y' — — y, r' — T. 



sin(-^)=^ = :=^ = -sin(9, 
r r 



cos (-(9)=^ = -= cos d, 
r r 



tan(-^) = ^ = ^:^ = -tand! 



Also, 



csc(— ^)= — csc^; sec (— ^)= seed; ctn (-- d)=— ctnd. 



VI, § 121] TRIGONOMETRIC FUNCTIONS 



177 



121. The Trigonometric Functions of 90° — 6. Figure 109 
represents angles 9 and 90° — 0, when ^ is in each of the f oui 
quadrants. Let OP be the terminal line of 6 and OP', the 



Y 


^ 


»' 

90'-9 

VF 





X' J 


^ X 




Fio. 109 

terminal line of 90° - 6. Take OP' = OP and let (x, y) be the 
coordinates of P and (a;', y') the coordinate of P. Then in all 
four j&gures we have 

^ = y> y'^^y r' = r. 
Hence 

sin(9O°-0) = 4=- = cos^, 
r r 



tan (90°-^) 



— ^ — IL — 
r r 

= ^ = ?=ctnd. 



cos(90°-^) = - = ^=sind, 
r r 



y 



Also, 



CSC (90° -^)=sec^, 
sec (90° -(9)=csc^; 
ctn(90° -^)=tand. 

Definition. The sine and cosine, the tangent and cotan- 
gent, the secant and cosecant, are called co-functions of each 
other. 

The above results may be stated as follows : Any function 
of an angle is equal to the corresponding co-function of the com- 
plementary angle.* 

*Two angles are said to be complementary if their sum is 90°, regardless of 
the size of the angles. 



178 MATHEMATICAL ANALYSIS [VI, § 122 

122. The Trigonometric Functions of 180° — 0. By draw- 
ing figures as in §§ 120, 121, the following relations may be 
proved : 

sin (180° - 0) = sin 0, esc (180° -6)= esc 0, 

cos (180° -0) = — cos e, sec (180° -6) = - sec 6, 

tan (180° - (9) = - tan 0, ctn (180° -&)=- ctn 6. 

The proof is left as an exercise. 

123. The Trigonometric Functions of 180° + 6. Similarly, 
the following relations hold : 

sin (180° + ^) = - sin 0, esc (180° + 0) = - esc 6, 

cos (180° 4- ^) = - cos 0, sec (180° + ^) = - sec 6, 

tan (180° + (9) = tan B, ctn (180° + ^) = ctn d. 

The proof is left as an exercise. 

124. Summary. An inspection of the results of §§ 120-123 
shows : 

1. Each function of — B or 180° ± B is equal in absolute value 
(but not always in sign) to the same function of 0, 

2. Each function of 90° — B is equal in magnitude and in sign 
to the corresponding co-function of 6. 

These principles enable us to find the value of any function 
of any angle in terms of a function of a positive acute angle 
(not greater than 45° if desired) as the following examples 
show. 

Example 1. Reduce cos 200° to a function of an angle less than 45°. 
Since 200° is in the second quadrant, cos 200° is negative. Hence 
cos 200°= - cos 20°. Why ? 

Example 2. Reduce tan 260° to a function of an angle less than 45°. 
Since 260° is in the third quadrant, tan 260° is positive. Hence 
tan 260° = tan 80° = ctn 10° (§ 121). 



VI, § 124] TRIGONOMETRIC FUNCTIONS 179 

EXERCISES 
Reduce to a function of an angle not greater than 45° : 

1. sin 163°. 6. esc 900°. 

2. cos(- 110°). 6. ctn (- 1215°). 
Ans. — cos 70° or — sin 20°. 7, tan 840°. 

3. sec (-265°). 8. sin 510°. 

4. tan 428°. 

Find without the use of tables the values of the following functions : 

9. cos 570°. 

13. cos 150°. 

10. sin 330°. 

11. tan 390°. W- tan 300°. 

12. sin 420°. 

Reduce the following to functions of positive acute angles : 
16. sin 250°. 18. sec (-245°). 

Ans. — sin 70° or — cos 20°. 19, esc (— 321°). 

16. cos 158°. 20. sin 269°. 

17. tan (-389°). 

21. Prove the following relations from a figure : 
(a) sin (90° + d) = cos d. (c) sin (180° + 6)= - sin $. 

cos (90° + e) = - sin d. cos (180° + ^) = - cos 0. 

tan (90° + d) = - ctn 0. tan (180° + 0) = tan 0. 

esc (90° + 0) = sec 0. CSC (180° + ^) = - esc ^. 

sec (90° + ^) = - CSC 0. sec (180° + 0) = -8ec0. 

Ctn (90° + 0)=- tan 0. ctn (180° -\-0) = ctn 0. 

(6) sin (180° -0)= sin 0. (d) sin (270° - ^) = - cos 0. 

cos (180° -0) = - cos 0. cos (270° -0) = - sin 0. 

tan ( 1 80° - ^) = - tan 0. tan (270° - ^) = ctn 0. 

CSC (180° - ^) = CSC 0. CSC (270° - ^) = - sec 0, 

sec (180° -0) = - sec 0. sec (270° -0) = - esc 0. 

ctn (180° - ^) = - ctn ^. ctn (270° - ^) = tan 0. 

(e) sin (270° + ^) = - cos ^. 
cos (270° + ^)= sin ^. 
tan (270° + 0) = — ctn 0. 
CSC (270° + ^) = - sec ^. 
sec (270° + 0)= CSC ^. 
ctn(270° + ^) = ~ tan^. 



180 



MATHEMATICAL ANALYSIS 



[VI, § 125 



125. Law of Sines. Consider any triangle ABC with the 
altitude CD drawn from the vertex C (Fig. 110). 




I) B D A 

Fia. 110 



In all cases we have sin A= -, sin B =- 

b 



a 



Therefore, dividing, we obtain 

sin^ a 



or 



sin^ 6' 
a b 



(2) 

sin A sin B 

If the perpendicular were dropped from B, the same argu- 
ment would give 

-^ = -^. (3) 

sin A sin C 

Combining results (2) and (3) we have 

a _ b _ c 

sin A sin 5 sin C* 

This law is known as the law of sines and may be stated as 

follows : 

Amj tivo sides of a triangle are proportional to the sines of the 

angles opposite these side'i. 

126. Law of Cosines. Consider any triangle ABC with the 
altitude CD drawn from the vertex C (Fig. 111). 
In Fig. Ill a 

AD = 6 cos ^ ; CD = 6 sin ^ ; DB = c — beosA. 
In Fig. Ill b 

AD = — 6 cos -4 ; CD = 6 sin ^ ; DB = c — b cos A 



VI, § 127] TRIGONOMETRIC FUNCTIONS 
In both figures 

Therefore 

a2 = c2 — 2 6c cos A-{-¥ cos^ A-\-ll^ sin^ A 
= c^ — 2 &c cos ^ 4- (cos2 A + sin* A) b% 



181 



b 




D B D 




whence 



FlQ. Ill 



a2 = &2 + c2 — 2 6c cos A, 



Similarly it may be shown that 

52 = c2 -f- a2 - 2 ca • cos B, 
c^ z= 0} ■\- h"^ — 2 ah • cos C. 

Any one of these similar results is called the law of cosines 
It may be stated as follows : 

Tlie square of any side of a triangle is equal to the sum of the 
squares of the other two sides diminished by tivice the product of 
these two sides times the cosine of their included angle.* 

127. Solution of Triangles. To solve a triangle is to find 
the parts not given, when certain parts are given. From 
geometry we know that a triangle is in general determined 
when three parts of the triangle, one of which is a side, 



* Of what three theorems in elementary geometry is this the equivalent ? 



182 



MATHEMATICAL ANALYSIS 



[VI, § 127 



are given.* Eight triangles have already been solved 
(§ 106 £f.), and we shall now make use of the laws of sines and 
cosines to solve oblique triangles. TJie methods employed 
will be illustrated by some examples. It will be found 
advantageous to construct the triangle to scale, for by so doing 
one can often detect errors which may have been made. 

128. Illustrative Examples. 

Example 1. Solve the triangle ABC, given 
^^^^^ A = 30^^ 20', B = 60° 45', a = 276. 

^A Solution : 

C = 180° - (^ + B)=1S0° - 91° 5' = 88° 55' ; 




a sin B 276 sin 60= 



sin J. 



sin 30= 20 



45^ ^ (276) (0.8725) ^^ygQ. 
I' 0.5050 ' ' 



also 



a sin G ^ 276 sin 38° 55' ^ (276) (0.i)998) ^ ^^g ^ 



Check : It is left as an exercise to show that for these values we have 
c2 = a2 + &2 _ 2 a6 cos C. 

Example 2. Solve the triangle ABC, given 
A = 30°, 6 = 10, a = 6. ^Q^ 

Constructing the triangle ABC, we see that .^x^so" 
two triangles AB\ C and AB2C answer the descrip- -^ 
tion since b>a> altitude CD. 

Solution : Now 




Fig. 113 



whence 
But 



-B2 



?H^ = ^orsin5i=^^Hl^ 
sin ^ a a 

Bi = 56.5°. 



180° - Bi = 180° - 56.5° = 123.6°, 



= 0.833, 



and 



Ci = 180° - (^ + Bi) = 180° - 86.5° = 93.6°, 
Ci = 180° -(A + Bi) = 180° - 153.5° = 26.6°. 

* When two sides and an angle opposite one of them are given, the triangle 
is not always determined. Why ? 



VI, § 128] TRIGONOMETRIC FUNCTIONS 183 

Now 



Also 



cg^ sin (72 ^ oj. c^^ a sin C^ ^ (6) (0.446) ^ g g^^ 
a sin ^ ' sin ^ 0.600 

Ci^ sinCi Qj, asm(7i^ (6)(0.998) ^^^Q3 

a sin^' sin J. 0.600 



Check : Cy^ = a^ -\- b^ — 2 ah cos (7i. 

143.5 = 86+ 100 +(2)(6)(10)(0.061) = 143.3. 

C22 = a2 4- 62 _ 2 a6 cos O2. 
28.62 = 36 + 100-(2)(6)(10)(0;896) - 28.60. 
Example 3. Solve the triangle ABC, given a= 10, 6=6, 0=40°. 
Solution : c^ = a^ + 6^ _ 2 a6 cos C j^ 

= 100 + 36 - ( 120) (0. 766) = 44.08. 
Therefore c = 6.64. Now \fy^ \V 

sin .4 = «^^5iZ = ^lOHO-643) ^ ^ ggg 
c 6.64 ' 

i.e. -4 = 104.6°. Likewise 

sin5 = ^«l5_^=(^KM43)^0.581, 
c 6.64 



Check : A + B^- G= 180.0°. 

Example 4. Solve the triangle ABC when 
a = 7, 6 = 3, c = 6. 
From the law of coshies, 

= -!=- 0.500, 





2 6c 2 

^ + c2-62 ^13^0Q^8, 
2ac 14 

^2 + 62 - C2 11 n 7QA 

COS C = = — = 0. 786. 

2 a6 14 

Therefore ^ ^ 120°, 5 = 21.8°, C = 38.2°. 

Check -. A^-B+ C= 160.0°. 



EXERCISES 

Solve the triangle ABC, given 
(a) ^ = 30°, 5 = 70°, a = 100; 

(6)^ = 40°, 5 = 70°, c = 110; 

(c) A = 45.5°, = 68.6°, 6 = 40 ; 
(d)J5 = 60.5°, C = 44°20', c = 20 ; 



184 MATHEMATICAL ANALYSIS [VI. § 128 

(e) a = 30, 6 = 64, 0=50°; (g)a=10, 6 = 12, c = U; 
(/) 6 = 8, a = 10, O = 60° ; {h) a = 21, 6 = 24, c = 28. 

2. Determine the number of solutions of the triangle ABO when 
(a) A = 30°, 6 = 100, a = 70 



(e) 


A = 


30°, 


6 = 


100, 


a = 


120 


(/) 


A = 


106°, 


6 = 


120, 


a = 


16 


ig) 


A = 


90°, 


6 = 


15, 


a = 


14. 



(6) A = 30°, 6 = 100, a = 100 

(c) ^ = 30°, 6 = 100, a = 50 

(d) A = 30°, 6 = 100, a = 40 

3. Solve the triangle ABC when 

(a) ^ = 37° 20', a = 20, 6 = 26 ; (c) ^ = 30°, a = 22, 6 = 34. 
(6) ^ = 37° 20', a = 40, 6 = 26; 

4. In order to find the distance from a point A to z. point JB, a line 
AC and the angles CAB and ACB were measured and found to be 
300 yd., 60° 30', 56° 10' respectively. Find the distance AB. 

5. In a parallelogram one side is 40 and one diagonal 90. The angle 
between the diagonals (opposite the side 40) is 25°. Find the length of 
the other diagonal and the other side. How many solutions ? 

6. Two observers 4 miles apart, facing each other, find that the angles 
of elevation of a balloon in the same vertical plane with themselves are 
60° and 40° respectively. Find the distance from the balloon to each- 
observer and the height of the balloon. 

7. Two stakes A and B are on opposite sides of a stream ; a third 
stake C is set 100 feet from A, and the angles ACB and CAB are observed 
to be 40° and 110°, respectively. How far is it from Ato B? 

8. The angle between the directions of two forces is 60°. One force 
is 10 pounds and the resultant of the two forces is 15 pounds. Find the 
other force.* 

9. Resolve a force of 90 pounds into two equal components whose 
directions make an angle of 60° with each other. 

10. An object B is wholly inaccessible and invisible from a certain 
point A. However, two points C and D on a line with A may be found 
such that from these points B is visible. If it is found that CD = 
800 feet, CA = 120 feet, angle DCB = 70°, angle CDB = 50°, find the 
length AB. 

* It is shown in physics that If the line segments AB 

3 ~ ^ ^ ^^^ -^^ represent in magnitude and direction two forces 

r ^^^/ acting at a point A, then the diagonal AJ) of the parallelo- 
fy^^ / gram ABCD represents both in magnitude and direction 

A B the resultant of the two given forces. 



VI, § 128] TRIGONOMETRIC FUNCTIONS 



185 



11. Given a, 6, A, in the triangle ABC. Show that the number of 
possible solutions are as follows : 

[ a < 6 sin ^ no solution, 
I bsmA<ia<.b two solutions, 



\a>b 
I a = 6 sin 



J 



one solution. 




( a <b no solution, 
I o > 6 one solution. 



12. The diagonals of a parallelogram are 14 and 16 and form an angle 
of 50°. Find the length of the sides. 

13. Resolve a force of magnitude 150 into two components of 100 and 
80 and find the angle between these components. 

14. It is sometimes desirable in surveying to extend a line such as AB 




in the adjoining figure. Show that this can be done by means of the 
broken line ABODE. What measurements are necessary ? 

15. Three circles of radii 2, 6, 5 are mutually tangent. Find the angles 
between their lines of centers. 

16. In order to find the distance between two objects A and B on op- 
posite sides of a house, a station C was chosen, and the distances CA 
= 500 ft., CB = 200 ft., together with the angle ACB - 65° SO' were 
measured. Find the distance from A to B. 

17. The sides of a field are 10, 8, and 12 
rods respectively. Find the angle opposite the 
longer side. 

18. From a tower 80 feet high, two objects, 
A and B, in the plane of the base are found to 
have angles of depression of 13° and 10° respec- 
tively ; the horiz(mtal angle subtended by A and B at the foot C of the 
tower is 44°. Find the distance from A to B. 




186 MATHEMATICAL ANALYSIS [VI, § 129 

129. Areas of Oblique Triangles. 
1. When two sides and the included angle are given. 
Denoting the area by S^ we have from geometry 
G S = ^ch, 

but ^ = & sin ^ ; therefore 
(4) S^^cb sin A. 

Likewise, 




Fig. 116 



S — ^ab sin O and S = ^acsinB. 



2. When a side and two adjacent angles are given. 

Suppose the side a and the adjacent angles B and C to be 
given. We have just seen that S = ^ac sin B. But from the 
law of sines we have 

a sin C 






Therefore 



S = 



SIR A 

a^ • sin jB • sin C 



2 sin A 

But sin A = sin [180° - (5 + (7)] = sin (5 + C). Therefore 
^ _ a'^ sin B sin O 
~ 2 sin (5+0)' 

3. When the three sides are given. 

We have seen that S = ^bc sin A. Squaring both sides of 
this formula and transforming, we have 

™ 7)2,.2 7)2^2 

S^ = ^ sin2^ = ^(l-cos2^) 



whence, 



= |(l4-cos^).|(l-cos^); 



^^A 1 b^-\-c^-a' 



S' = ^ 1-h 



2 V 2 6c 



;N bcf. b^ + c^-a^ \ 
J 2[ 2 be J 



VI, § 130] TRIGONOMETRIC FUNCTIONS 187 

4 ' 4 

_M-_c4;_a b -\-c—a a~b-\-c g-f 6~c 
2*2*2*2' 

which may be written in the form 

S^ = s(s-a){s-b)(s-c), 
where 2s = a + &4-c. Therefore, 



(5) S = Vs(s-a)(s-6)(s-c). 

130. The Radius of the Inscribed Circle. If r is the radius 
of the inscribed circle, we have from elementary geometry, 
since s is half the perimeter of the triangle, S = rs ; equating 
this value of S to that found in equation (5) of the last article 
and then solving for r, we get, 



^J (s-a)(s-b)(s-c) _ 

^ 5 



EXERCISES 
Find the area of the triangle ABC, given 

1. a = 26, b = 31.4, C = 80° 25'. 4. a = 10, 6 = 7, = 60°. 

2. 6 = 24, c = 34 3, ^ = 60° 25'. 5. a = 10, 6 = 12, C=60°. 

3. a = 37, 6 = 13, C = 40°. 6. a = 10, 6 = 12, C = 8°. 

7. Find the area of a parallelogram in terms of two adjacent sides 
and the included angle. 

8. The base of an isosceles triangle is 20 ft. and the area is 100/ \/3 
sq. ft. Find the angles of the triangle. Ans. 30°, 30°, 120°. 

9. Find the radius of the inscribed circle of the triangle whose sides 
are 12, 10, 8. 

10. How many acres are there in a triangular field having one of its 
sides 60 rods in length and the two adjacent angles, respectively, 70° 
and 60° ? 



CHAPTER VII 

TRIGONOMETRIC RELATIONS 

131. Radian Measure. In certain kinds of work it is more 
convenient in measuring angles to use, instead of the degree, 
a unit called the radian. A radian is defined as the angle at 
the center of a circle whose subtended arc is equal in length 
to the radius of the circle (Fig. 117). Therefore, if an angle 6 
at the center of a circle of radius r units subtends an arc of 
s units, the measure of in radians is 

(1) 0=i. 

Since the length of the whole circle is 2 irr, it follows that 

^^ = 2 TT radians = 360°, 
r 




(2) IT radians = 180°. 
Fig. 117 Therefore, 

-IQAO 

1 radian = ±^ = 57° 17' 45" (approximately). 

TT 

It is important to note that the radian * as defined is a con- 
stant angle, i.e., it is the same for all circles, and can therefore 
be used as a unit of measure. 

* The symbol »" is often used to denote radians. Thus 2" stands for 2 
radians, tt*- for tt radians, etc. When the angle is expressed in terms of -t (the 
radian being the unit) , it is customary to omit »•. Thus, when we refer to an 
angle tt, we mean an angle of tt radians. When the word radian is omitted, it 
should be mentally supplied in order to avoid the error of supposing ir means 
180. Here, as in geometry, IT = ;i. 14159. ... 

188 




VII, § 132] TRIGONOMETRIC RELATIONS 189 

Erom relation (2) it follows that to convert radians into 
degrees it is only necessary to multiply the number of radians 
by ISO/V, while to convert degrees into radians we multiply 
the number of degrees by 7r/180. Thus 45° is 7r/4 radians ; 
7r/2 radians is 90°. 

132. The Length of Arc of a Circle. From relation (1), 
§ 131, it follows that s=re 

5 = re. 

That is (Fig. 118), if a central angle is measured 
in radians, and if its intercepted arc and the 
radius of the circle are measured in terms of 
the same unit, then ^°' 

length of arc = radius x central angle in radians. 

EXERCISES 

1. Express the following angles in radians : 

25°, 145°, 225°, 300°, 270°, 450°, 1150°. 

2. Express in degrees the following angles : 

IT 7 IT Sir n 57r 

4' "T' T'^'^'T* 

t. A circle has a radius of 20 inches. How many radians are there in 
an angle at the center subtended by an arc of 25 inches ? How many 
degrees are there in this same angle ? Ans. f ; 71° 37' approx. 

4. Find the radius of a circle in which an arc 12 inches long subtends 
an angle of 35°. 

6. The minute hand of a clock is 4 feet long. How far does its ex- 
tremity move in 22 minutes ? - 

6. In how many hours is a point on the equator carried by the rotation 
of the earth on its axis through a distance equal to the diameter of the earth ? 

7. A train is traveling at the rate of 10 miles per hour .on a curve of 
half a mile radius. Through what angle has it turned in one minute ? 

8. A wheel 10 inches in diameter is belted to a wheel 3 inches in 
diameter. If the first wheel rotates at the rate of 5 revolutions per 
minute, at what rate is the second rotating ? How fast must the former 
rotate in order to produce 6000 revolutions per minute in the latter ? 



190 MATHEMATICAL ANALYSIS [VII, § 133 

133. Angular Measurement in Artillery Service. The 

divided circles by means of which the guns of the United States Field 
Artillery are aimed are graduated neither in degrees nor in radians, but 
in units called mils. The mil is defined as an angle subtended by an arc 
of ■g:^^ of the circumference, and is therefore equal to 

2_ir_ ^ SAAW ^ 0.00098175 = (0.001 - 0.00001825) radian. 
6400 3200 ^ 

The mil is therefore approximately one thousandth of a radian. 
(Hence its name.)* 
Since (§132) 

length of arc = radius x central angle in radians, 
it follows that we have approximately 

T5) li ill A 

length of arc = x central angle in mils ; 

1000 ^ ' 

i.e. length of arc in yards = (radius in thousands of yards) • (angle 
in mils). The error here is about 2 %. 

Example 1. A battery occupies a front of 60 yd. If it is at 
5500 yd. range, what angle does it subtend (Fig. 119)? We 
have, evidently, 

angle = — = 11 mils. 
5.5 

Example 2. Indirect Fire, t A battery posted with its right gun at G 
is to open fire on a battery at a point T, distant 2000 yd. and invisible 

* To give an idea of the value in mils of certain angles the following has 
been taken from the Drill Regulations for Field Artillery (1911), p. 164: 

" Hold the hand vertically, palm outward, arm fully extended to the front. 
Then the angle subtended by the 

width of thumb is 40 mils 

width of first finger at second joint is 40 mils 

width of second finger at second joint is .... 40 mils 

width of third finger at second joint is 35 mils 

width of little finger at second joint is 30 mils 

width of first, second, and third fingers at second joint is . 115 mils 
These are average values." 

t The limits of this text preclude giving more than a single illustration of 
the problems arising in artillery practice. For other problems the student is 
referred to the Drill Regulations for Field Artillery (1911), pp. 57, 61, 150-164; 
and to Andrews, Fundamentals of Military Service, pp. 153-159, from which 
latter text the above example is taken. 




VII, § 133] TRIGONOMETRIC RELATIONS 



191 



from Cr (Fig. 120) . The officer directing the fire takes post at a point 
B from which both the target T ami a church spire P, distant 3000 yd. 
from O are visible. B is 100 yd. at the right of the line &T and 120 yd. 
at the right of the line GP and the officer finds by measurement that the 
angle P-BT contains 3145 mils. In order to train the gun on the target 
the gunner must set off the angle PGT on jf rpr 

the sight of the piece and then move the gun 
until the spire P is visible through the sight. 
When this is effected, the gun is aimed at T. 

Let F and E be the feet of the perpen- 
diculars from B to G^Tand G^P respectively, 
and let BV and BP' be the parallels to 
OT and OP that pass through B. Then, 
evidently, if the officer at B measures the 
angle PPT, which would be used instead 
of angle POT were the gun at B instead 
of at G^ and determines the angles TBT' = 
FTB and PBP' = EPB, he can find the 
angle PGT from the relation 




PGT=P'BT'= PBT 



Fig. 120 
TBT'- PBP'. 



Now 



tan FTB 



FB 

TF 



tan EPB = — . 
PF 



Furthermore if FTB and EPB are small angles, i.e., if FB and EB are 
small compared with OT Sind OP respectively, the radian measure of the 
angle is approximately equal to the tangent of the angle. Why ? Hence 

we have 

FB 
GT 
EB 
OP 
100 



Therefore 



FTB - tan FTB = 
EPB = tan EPB 



approximately. 



TBT' = FTB = -^^^ radians 
2000 



50 mils. 



PBP' 



EPB = -^ radians 
3000 



40 mils. 



Hence POT = PBT - TBT - PBP* 

= 3145 - 60 - 40 
— 3056 mils, 

which is the angle to be set off on the sight of the gun. 



1^2 MATHEMATICAL ANALYSIS [VII, § 133 

Hence for the situation indicated in Fig. 120 we have the following 
rule : * 

(]) Measure in mils the angle PBT from the aiming point P to the 
target T as seen at B. 

■ : (2) Measure or estimate the offsets FB and EB in yards, the range 
G^Tand the distance GB of the aiming point P in thousands of yards. 

(3) Compute in mils the offset angles by means of the relations 

TBT = FTB, 
FBP' = EPB, 

TBr=—, 
GT' 

CrF 

(4) Then the angle of deflection FGT is equal to the angle FBT 
diminished by the sum of the offset angles. 

EXERCISES 

1. A battery occupies a front of 80 yd. It is at 5000 yd. range. 
What angle does it subtend ? 

2. In Fig. 120 suppose FBT= 3000 mils, FB = 200 yd., GT = 3000 yd., 
EB = 150 yd., GP = 4000 yd. Find the number of mils in FGT. 

3. A battery at a point G is ordered to take a masked position and be 
ready to fire on an indicated hostile battery at a point T whose range is 
known to be 2100 yd. The battery commander finds an observing station 
B, 200 yd. at the right and on the prolongation of the battery front, and 
175 yd. at the right of PGr An aiming point P, 5900 yd. in the rear, is 
found, and PBT is found to be 2600 mils. Find FGT. 

134. Inverse Trigonometric Functions. The equation 

X = sin y (1) 

may be read : 

y is an angle whose sine is equal to a;, 

a statement which is usually written in the contracted, form 

y — arc sinflj.f (2) 

* There are three cases with corresponding rules, depending on whether P 
is in front of, rear of, or on the flank of G. 

t Sometimes written y = sin-i.c. Hefe — 1 is not an algebraic exponent, 
but merely a part of a functional symbol. When we wish to raise sin x to 
the power — 1, we write (sin a;)-i. 



VII, § 134] TRIGONOMETRIC RELATIONS 



193 



For example, x = sin 30° means that x = ^, while y = arc sin^ 
means that y = 30°, 150°, or in general (n being an integer), 

30° + n . 360° ; 150° + n • 360°. 

Since the sine is never greater than 1 and never less than 
— 1, it follows that — 1 ^ a; ^ 1. It is evident that there is 
an unlimited number of values of y = arc sin x for a given value 
of X in this interval. 

We shall now define the principal value Arc sin x* of arc sin x^ 

distinguished from arc sin x by the use of the capital A, to 

be the numerically smallest angle whose sine is equal to x. 

This function like arc sin x is defined only for those values of 

X for whichsk 

- 1< a; < 1. 



The difference between arc sin x and Arc sin x is well illus- 
trated by means of their graph. It is 
evident that the graph oi y = arc sin x, 
i.e. X = sin y is simply the sine curve 
with the role of the x and y axes inter- 
changed. (See Fig. 121.) Then for every 
admissible value of x, there is an un- 
limited number of values of y ; namely, 
the ordinates of all the points Pi, P2, — , in 
which a line at a distance x and parallel 
to the 2/-axis intersects the curve. The 
single-valued function Arc sin x is repre- 
sented by the part of the graph between 
Jlf and iV. 

Similarly arc cos x, defined as " an angle whose cosine is a;," 

* Sometimes written Sin-icc, distinguished from sin-^a; by the ase of the 
capital S. 





T 


-. 








2t 


^ 


Ps 






( sir 










ir 


\ 


n 






IT 


i 


> 

y 


N- 


-i 


^ 






1 X 


M 


G 










y= arc sin x 






y= 


'Ar 


(;sinx 



Fig. 121 



194 



MATHEMATICAL ANALYSIS [VII, § 134 



has an ■anlimited. number of values for 
every admissible value of a; (— 1^ a; ^ 1). 
We shall define the principal value Arc 
cos X as the smallest positive angle whose 
cosine is x. That is, 

< Arc cos a? ^ TT. 

Figure 122 represents the graph of 
y = arc cos x and the portion of this graph 
between M and N represents Arc cos x. 

Similarly we write x = tan y as y,= arc 
tan X, and in the same way we define the 
symbols arc ctn x ; arc sec x ; arc esc x. 

The principal values of all the inverse trigonometric functions 

are given in the following table. 





7 










2ir 




■) 






3jr 




Ps 




N 


^■J 










IT 


^ 


\ 


M 


-1 




IT 




< 


1 X 



y= arc cos x 
y=Arc cosz 
Fig. 122 



y = 


Arc sin x 


Arc cos X 


Arc tan x 


Range of x 
Range of y 

X positive 
X negative 


-^to^ 
2 2 

1st Quad. 

4th Quad. 


-l^x^l 
to T 

1st Quad. 
2d Quad. 


all real values 

1st Quad. 
4th Quad. 




Arc ctn x 


Arc sec x 


Arc CSC X 


Range of x 
Range of y 

X positive 
X negative 


all values 

tOTT 

1st Quad. 
2d Quad. 


a; > 1 or a; < - 1 

tOTT 

1st Quad. 
2d Quad. 


X>l0TX<-l 

-^ to"^ 
2 2 

1st Quad. 

4th Quad. 



In so far as is possible we select the principal value of each 
inverse function, and its range, so that the function is single- 
valued, continuous, and takes on all possible values. This ob- 
viously cannot be done for the Arc sec x and for Arc esc y. 



VII, § 134] TRIGONOMETRIC RELATIONS 195 

EXERCISES 

1. Explain the difference between arc sin x and Arc sin x. 

2. Find the values of the following expressions : 

(a) Arc sin \. (&) arc sin \. (c) arc tan 1. 

(d) Arc tan - 1. (e) arc cos I^. (H Arc cos lA. 

2 ' 2 

3. What is meant by the angle tt ? 7r/4? 

4. Through how many radians does the minute hand of a watch turn 
in 30 minutes ? in one hour ? in one and one half hours ? 

5. For what values of x are the following functions defined : 

(a) arc sin x ? (6) arc cos cc? (c) arc tan x ? 

id) arc ctn X ? (e) arc sec a; ? (/) arc esc cc ? 

6. What is the range of values of the functions : 

(a) Arc sin x ? (&) Arc cos x ? (c) Arc tan x ? 

id) Arc ctn x ? (e) Arc sec x ? (/) Arc esc x ? 

7. Draw the graph of the functions : 

(a) arc sin x. (&) arc cos x. (c) arc tan x. 

(<?) arc ctn X. (e) arc sec x. (/) arc esc x. 

8. Find the value of cos (Arc tan |). 

Hint. Let Arc tan \ = 6. Then tan ^ = f and we wish to find the 
value of cos e. 

9. Find the values of cos (arc tan |). 

10. Find the value of the following expressions : 

(a) sin (arc cos |). (c) cos (Arc cos y\). (e) sin (Arc sin \). 

(&) sin (arc sec 3). {d) sec (Arc esc 2). (/) tan (Arc tan 6). 

11. Prove that Arc sin (2/6) = Arc tan (2/V2T). 

12. Find x when Arc cos (2 x^ — 2 x) = 2 7r/3. 

Find the values of the following expressions : 

13. cos [90^ — Arc tan |]. 

14. sec [90° -Arc sec 2]. 

15. tan [90° - Arc sin ^{\. 



196 



MATHEMATICAL ANALYSIS [VII, § 135 



135. Projection. Consider two directed lines p and g in a 
plane, i.e. two lines on each of which one of the directions 
has been specified as positive (Fig. 123). Let A and B be 
any two points on p and let A\ B' be the points in which per- 




Fig. 123 

pendiculars to q through A and B, respectively, meet q. The 
directed segment A'B' is called the projection of the directed seg- 
ment AB on q and is denoted by 

A'B' = proj^ AB. 

In both figures AB is positive. In the first figure A'B' is posi- 
tive, while in the second figure it is negative. 

As special cases of this definition we note the following : 

1. If p and q are parallel and are directed in the same way, 

we have 

^m]^AB=:AB. 

2. If p and q are parallel and are directed oppositely, we 

have 

^xo]^AB=-AB. 

3. If p is perpendicular to q, we have 

proj,^B = 0. 

It should be noted carefully that these propositions are true 
no matter how A and B are situated on p. 

We may now prove the following important proposition; 



VII, § 135] TRIGONOMETRIC RELATIONS 



197 



If Aj B are any two points on a directed line p, and q is 
any directed line in the same plane with p, then we have both in 
Tnagnitvde and sign : 

(1) proj\ AB = AB' cos {qp), 

where (qp) represents an angle through ichich q must be rotated 
in order to make its direction coincide with the-direction of p. 

We note first that all possible determinations of the angle 
(qp) have the same cosine, since any two of these determina- 
tions differ by multiples of 360° (Fig. 124). We shall prove 




Fig. 124 
the proposition first for the case where AB has the same direc- 
tion as p, i.e. where AB is positive. To this end we draw 
through A (Fig. 125) a line qi parallel to q and directed in the 



A' 



\Z. 




B' 



^ 


A Bi 






^ 


B 




J. 


I' J 


S' ' 


^ 



Fig. 125 
same way. (We may evidently assume without loss of gener- 
ality that q is horizontal and is directed to the right.) 

Let A'B' have the same significance as before and let BB' 
meet ^'i in Bi. Then, by the definition of the cosine, we have 

A 7? 

—^ = cos (q^p) = cos (qp)y 

in magnitude and in sign ; or 

ABi = ^5 cos (qp). 



But 
Therefore 



AB, = A'B'=VTOj^AB. 
projg AB = AB cos (qp). 



198 MATHEMATICAL ANALYSIS [VII, § 135 

Finally, if AB is negative, BA is positive, and, by the result 
just obtained, we should have 

B'A' == BA cos (qp). 

Hence, changing signs on both sides of this equation, we 
have 

A'B' = AB cos (q 2^). 

The special cases 1, 2, 3 listed on p. 196 are obtained from 
formula (1) by placing (qp) equal to 0°, 180°, 90°, respectively ; 
for cos 0° = 1, cos 180° - - 1, cos 90° = 0. 

136. Application of Projection. In Physics, forces and 
velocities are usually represented by line segments. A force 
of 20 pounds, for example, is represented by a segment 20 units 
in length and drawn in the direction of the force. A velocity 
of 20 feet per second is represented by a segment 20 units in 
length and drawn in the direction of the motion. 

The projection on a given line I of a segment representing 
a force or velocity represents the component of the force or 
velocity in the direction of l. 

Example. A smooth block is sliding down a smooth incline 
which makes an angle of 30° with the horizontal. If the block 
weighs 10 lb., what force acting directly up 
the plane will keep the block at rest ? 

Draw the segment AB 10 units in length, 

directly downward to represent the force 

exerted by the weight. Project this segment 

Fig. 126 ^^ ^^^ incline and call this projection AC. 

Now angle ABC = 30°. Therefore AC = AB sin 30° = 5. This 

is the component of the force AB down the plane. Therefore 

a force of 5 lb. acting up the plane will keep the body at rest. 




A 




•^ 


B 


a! c" 


B' 



VII, § 136] TRIGONOMETRIC RELATIONS 199 

Theorem. If A, B, C are any three points in a plarie, arid I 
is any directed line iii the plane, the algebraic sum of the projec- 
tions of the segments AB and BC on I is equal to the projection of 
the segment AC on I. 

As a point traces out the path, from A to B, and then from 
B to O (Fig. 127), the projection of the poiiit traces out the 
segments from A' to B' and then from B' 
to C. The net result of this motion is a 
motion from A' to C which represents 
the projection of AC, i.e. 

A'B' + B'C = A'C, Fig. 127 

EXERCISES 

1. What is the projection of a line segment upon a line I, if the line 
segment is perpendicular to the line I ? 

2. Find proj^^ AB and proj^ AB* in each of the following cases, if a 
denotes the angle from the cc-axis to AB. 

(a) AB = 5, a = 60°. (c) AB = 6, a = 90°. 

(6) AB = 10, a = 300°. (d) AB = 20, a = 210°. 

3. Prove hy means of projection that in a triangle ABC 

a = b cos C + c cos B. 

4. If proja; AB = 3 and projy AB = — 4, find the length of AB. 

6. A steamer is going northeast 20 miles per hour. How fast is it 
going north ? going east ? 

6. A 20 lb. block is sliding down a 15° incline. Find what force 
acting directly up the plane will just hold the block, allowing one half a 
pound for friction. 

7. Prove that if the sides of a polygon are projected in order upon any 
given line, the sum of these projections is zero. 

* Proji AB and projy AB mean the projections of AB on the x-axis and 
the y-axis, respectively. 



200 



MATHEMATICAL ANALYSIS [VII, § 137 



137. Rotation in a Plane. Suppose that a point P{Xj y) in 
a plane moves on the arc of a circle with center at the origin 0, 
through an angle a. Suppose that its position after this 
rotation is Pix^ y') referred to the same axes of coordinates. 
We desire to find x' and y' in terms of a;, y, and a. 

In Fig. 128 we have 
drawn P and its coordi- 
nates X = OM, y = MP, and 
the new position OM' P' of 
the triangle OMP after a 
rotation about the origin 
through an angle a. The 
coordinates x' = ON, ?/' = 
NP' of P are the pro- 
jections of OP on the 
a;-axis and the y-axis re- 
spectively, and these pro- 
jections are equal respec- 
tively to the sum of the projections of OM' and M'P on the 
respective axes. Hence, 

x' = proj, OP = proj, OM' -f- proj, M'P 

= OM' cos {OX, OM') 4- M'P cos (OX, M'P) 

= X cos a -\- y cos (a -1- 7r/2) 

= X cos a — y sin a. 
y' = proj, OP = proj^ OM' -f- proj^ M'P' 

= OM' cos (OF, 03/') -f 3/'P' cos ( OY, M'P') 

= X cos (— 7r/2 + «) -f- 2/ cos a 

= X sin a -h y cos a. 

Therefore, if the point P{x, y) is rotated about the origin 
through an angle a, the coordinates (x', y') of its new position 
are given by the formulas 



Y> 


\ 


1 








P'l 
1 ^ 


N. CC 


m 






k 






L / 




P 




hs\^ 




V 




'/>'''x' \ 









1^ 


T X 




¥ ^X 



Fig. 128 



VII, § 138] TRIGONOMETRIC RELATIONS 



201 



(1) 



J x' = X cos a — y sin a 
[j/ = x sin a + 1/ cos a. 

It should be noted that the above method of derivation is 
entirely general, i.e. it will apply to a point P in any quad- 
rant and to any angle a. 

138. The Addition Formulas. We may now enter upon a 
more detailed study of the properties 
of the trigonometric functions. We 
shall first express sin (a + p) and 
cos (a + /8) in terms of sin a, cos a, 
sin p, cos 13.* To this end let OP be 
the terminal side of any angle a (Fig. 
129). If OP is then rotated about 
through an angle jS to the position 
OP, the terminal line of the angle 
a -\- p is OP'. If P has the coordinates (x, y) and P the 
coordinates {x\ y'), then from (1) § 137, 

x' = xcos p — y sin /8, 
y' = X sin ^ + y cos p. 
Now sin {a + y8) is by definition equal to ^ and cos (a 4- /?) 

OP' = OP. Hence 

sin(a+ fi)=^ = - sin/3 4-^ cos^S, 
7- r r 

sin (a + P) = sin a cos p + cos a sin p. 



T 


/ 


Mv) 






/ 


'' p^ 




L^ 




V 





W" 


X 


X 



Fig. 129 



to — where r 
r 



or 

(1) 



Also 



or 
(2) 



cos (a + ;8) = - = - cos ^ - ^ sin^, 
r r r 

cos (a 4- P) = cos a cos p — sin a sin p. 



♦We have already had occasion to note that sin (a + ^) is not in general 
equal to sin a + sin ^. (See Ex. 5, p. 151.) 



202 MATHEMATICAL ANALYSIS [VII, § 138 

Further we have 

tan (a4- B) = sin (a -\- /?) __ sin « cos ^ + cos ce sin yg 
cos {a -f- y8) cos a cos /3 — sin a sin)8* 

Dividing numerator and denominator by cos a cos )8, we have 

(3) tan(a + 3) = ^^ + ^^ °^. 

Furthermore, by replacing yg by — /? in (1), (2), and (3), and 
recalling that 

sin (—/?)= — sin 13, cos (— ^)= cos /?, tan (— yS) = — tanyS, 

we obtain 

(4) sin(a— p)=sinacosp— cosasinp, 

(5) cos (a — p) = cos a cos p + sin a sin p, 

(6) tanCa- S)= tang-tanp 
v; i^ii^a p; i_^tanatanp 



EXERCISES 

Expand the following : 

1. sin (45° + a)= 3. cos (60'' + a) = 5. sin (30° - 45°) = 

2. tan (30°-^)= 4. tan (45° + 60°) = 6. cos ( 180° - 45°) = 

7. What do the following formulas become if a = yS ? 

sin (a + /3) = sin a cos ^3 -f cos a sin /3. ^^^ /^ , on _ tan ct + tan j3 

sin (a — /3) = sin a cos j3 — cos a sin jS. ' ~ 1 - tan a tan /3 * 

cos (a + /3) = cos a cos /3 - sin a sin ^. . , „. _ tan a — tan/3 

lan fcc — p) — ■ ■ — — — — . 

cos (a — /3) = cos a cos ^ + sm a sin /3. 1 + tan a tan )3 

8. Complete the following formulas : 

sin 2 a cos a + cos 2 a sin a = tan 2 « + tan a _ 

sin 3 a cos a — cos 3 a sin a = 1 — tan 2 a tan a ~ 

9. Prove sin 75° =: ^^ + \ cos75°=^^-:\ tan75° = ^5+^. 

2\/2 2V2 V3-1 

10. Given tan a = |, sin /3 = ^\, and « and ^ both positive acute angles, 
find the value of tan (a -f /3) ; sin (a — /3) ; cos(a + /3) ; tan (a - /3). 



VII, § 138] TRIGONOMETRIC RELATIONS 203 

11. Prove that 

(a) cos (60° + a) + sin (80° + «) = cos a. 

(b) sin (60° + 6)- sin (60° - ^) = sin d. 

(c) cos (30° -f ^) - cos (30° - d) = -^in d. 

(d) cos (45° + d)+ cos (45° -6) = V2 • cos 6. 

(e) sin(a + -J + sinfa — -j = sina. 



(/) cos(a + ^) + cos(a-^) = V3. 



(g) tan (45° + ^) = L±ia!L? . (h) tan (46° - 5) = ^ ^ ^^° ^ • 

12. By using the functions of 60° and 30° find the value of sin 90° ; 
cos 90°. 

13. Find in radical form the value of sin 15°; cos 15°; tan 16° J 
sin 105° ; cos 105° ; tan 105°. 

14. If tan a = I, sin j3 = j\, and a is in the third quadrant while /3 is 
in the second, find sin (a ± j3) ; cos (a ± /3) ; tan (a ± /3). 

Prove the following identities : 

j^g sin (ct + /3) _ tan ct + tan /3 jg sin2cs ^^^ ^ ^ = ejn 3 « 

sin (a — j3) tan a — tan /3 ' sec a esc a 

j^ tan ot — tan (cc — /3) _ ^^ o 19- (a) sin (180° — &) = sin ^. 

l + tanatan(a -^) ~ (&) cos(180° - ^) =- cos^. 

18. tan(0 ± 45°) + ctn {6 T 45°) = 0. (c) tan (180° - ^) = - tan Q. 

20. cos (a + /3) cos (a — /3) = cos^ a — sin^ ^. 

21. sin (a + jS) sin (a — /3) = sin2 a — sin2 /3. 

22. ctn(«+^) = ^^"""^'^^-^ 23. ctn («-^) = ^ill^^^^+i . 

ctn a + ctn j3 ctn /3 - ctn a 

24. Prove Arc tan ^ + Arc tan \ = ir/i 

[Hint : Let Arc tan l = x and Arc tan 1 = y. Then we wish to prove 
X -i-y = 7r/4, which is true since tan (x + y)= 1.] 

25. Prove Arc sin a + Arc cos a = - , if < a < 1. 

26. Prove Arc sin j\ + Arc sin f = Arc sin ||. 

27. Prove Arc tan 2 -}- Arc tan ^ = 7r/2. 

28. Prove Arc cos | + Arc cos (— ^j) = Arc cos ( — f f ). 

29. Prove Arc tan j% + Arc tan | = Arc tan ||. 

30. Find the value of sin [Arc sin | + Arc ctn |]. 

31. Find the value of sin [Arc sin a -H Arc sin 6] if < a < 1, < 6 < 1. 



204 MATHEMATICAL ANALYSIS [VII, § 138 

32. Expand sin (x -{ y -\- z) ; cos(x + y + z). 
[Hint: a; + 2/ + 2 = (x + ?/) + 2.] 

33. The area ^ of a triangle was computed from the formula 
A = I ab sin 6. If an error e was made in measuring the angle ^, show that 
the corrected area A' is given by the relation A' = A{co8€ + sinectn^). 

139. Functions of Double Angles. In this and the follow- 
ing articles (§§ 139-141) we shall derive from the addition 
formulas a variety of other relations which are serviceable in 
transforming trigonometric expressions. Since the formulas 
for sin (a + /3) and cos (a + yS) are true for all angles a and fi, 
they will be true when JS = a. Putting /? = a, we obtain 

(1) sin 2 a = 2 sin a cos a, 

(2) cos 2 a = cos2 a — sin2 a. 
Since sin^ a + cos^ a = 1, we have also 

(3) cos 2 a = 1 - 2 sin2 a 

(4) = 2 C0S2 a — 1. 

Similarly the formula for tan (a + /?) (which is true for all 
angles a, ^, and a + /3 which have tangents) becomes, when 
)8 = «, 

^ ^ 1 — tan^a 

which holds for every angle for which both members are 
defined. 

The above formulas should be learned in words. For ex- 
ample, formula (1) states that the sine of any angle equals 
twice the sine of half the angle times the cosine of half the 

angle. Thus 

sin 6 a; = 2 sin 3 x cos 3 a?, 

2 tan 2 x 



tan 4 x = 



l~tan22a;' 



cos a; =5 cos*?— sin^?. 



VII, § 140] TRIGONOMETRIC RELATIONS 205 

140. Functions of Half Angles. From (3), § 139, we have 





2sin^^ = 


; 1 — COS a. 


Therefore 


• «-!= 




(6) 


:±v'~r" 


From (4), 


§ 139, we have 




Therefore 


2cos2^ = 
cos| = 


1 + COS a. 


(J) 


±V'T"- 



rormulas (6) and (7) are at once seen to hold for all angles 
a. Now, if we divide formula (6) by formula (7), we obtain 



(8) ta"i=±Vr 



— cos a 



-j- cos a 

which is true for all angles a except n • 180°, where n is any 
odd integer. 

Example. Given sin ^ = — 3/5, cos A negative ; find sin (A/2). 

Since the angle A is in the third quadrant, A/2 is in the second or 
fourth quadrant, and hence sin (A/2) may be either positive or negative. 
Therefore, since cos A=— 4/5, we have 

2 Al 2 VlO 10 

EXERCISES 

Complete the following formulas and state whether they are true for 

all angles : 

1. sin 2 a = 3. tan 2 a = 6. cos ^ = 

2 

2. cos2a= (three forms). 4. sin-i= 6. tan- = 

2 ' ^ 

7. In what quadrant is e/2 if 6 is positive, less than 360°, and in the 
second quadrant ? third quadrant ? fourth quadrant ? 



206 MATHEMATICAL ANALYSIS [VII, § 140 

8. Express cos 2 a in terms of cos 4 a. 

9. Express sin x in terms of functions of 3 x. 

10. Express tan 4 a in terms of tan 2 a. 

11. Express tan 4 a in terms of cos 8 a. 

12. Express sin x in terms of functions of x/2. 

13. Explain why the formulas for sin x and cos x in terms of functions 
ot2x have a double sign. 

14 From the functions of 30° find those of 60°. 

15. From the functions of 60° find those of 30°. 

16. From the functions of 30° find those of 16°. 

17. From the functions of 15° find those of 7.5°. 

18 Find the functions of 2 a if sin a = f and a is in the second 
quadrant. 

19. Find the functions of a/2 if cos a =— 0.6 and a is in third quad- 
rant, positive, and less than 360°. 

20. Express sin 3 a in terms of sin a. [Hint : 3a = 2a-f-a.] 

21. From the value of cos 45° find the functions of 22.5". 

22. Given sin a = — and a in the second quadrant. Find the values of 

13 ^ 

(a) sin 2 a. (c) cos 2 a. (e) tan 2 a. 

(b) sin". (d) cos-. (/) tan-. 

2 2 2 

o 

23. If tan 2 a = - find sin a, cos a, tan a if a is an angle in the third 

4 
quadrant. 

Prove the following identities : 

24. L±-^2S^=ctn«. 27. 1 - cos 2 ^ + sin 2 g^ ^^^ ^ 

sin a 2 1 -I- cos 2 ^ 4- sin 2 ^ 

26. fsin^-cos^j = 1 - sin 5. 28. sin^ + cos^ = ± Vl + sin«. 

26. cos2g + cosg-H^^^^g 29. sec« + tan a = tanf? + «^ 

siu2e + sin^ \4 2/ 

30. 2 Arc cos x = Arc cos (2 a;* — 1) . 



31. 2 Arc coso; = Arcsin (2 arvl — a;2). 



VII, § 141] TRIGONOMETRIC RELATIONS 207 



32. tan [2 Arc tan x] = -^-^ . 34. tan [2 Arc sec x] = ± ^ ^ 



1-x-^ - 2 - x-^ 

1 — r2 
33. cos [2 Arc tan x] = ~ — • 35. cos (2 Arc sin a) = 1 - 2 a^. 

•■ -• 1 + x^ 

Solve the following equations : 

36. cos 2 X + 5 sin X = 3. 40. sin^ 2 x — sin^ x = |. 

37. cos 2 X - sin X = |. 41. sin 2 x = 2 cos x. 

38. sin 2 X cos X = sin X. 42. 2 sin22x = 1 — cos2x. 

39. 2 sin2 x + sin^ 2 x = 2. 43 ctn x — esc 2 x = 1. 

44. A flagpole 50 ft. high stands on a tower 49 ft. high. At what dis- 
tance from the foot of tlie tower will the flagpole and the tower subtend 
equal angles ? 

45. The dial of a town clock has a diameter of 10 ft. and its center is 
100 ft. above the ground. At what distance from the foot of the tower 
will the dial be most plainly visible ? [The angle subtended by the dial 
must be as large as possible.] 

141. Product Formulas. From § 138 we have 

sin (a 4- j8) = sin a cos ft -\- cos a sin p, 
sin (a — ^) = sin a cos ^ — cos a sin y8. 
Adding, we get 

(1) sin (a -f iS) + sin {a — 13)= 2 sin a cos jS. 
Subtracting, we have 

(2) sin (a -f /3) — sin (« — ^) = 2 cos a sin ft. 
Now, if we let a + /8 = P and a — fi = Q, 

then « = ^«, fi = ^. 

Therefore formulas (1) and (2) become 

. P 4- O P — O 
sm P H- sm Q = 2 sm — :r_J!^cos — — -^, 
2 2 

P 4- O . P— O 
Sin P — sm Q = 2 cos ^ ^ sm — — ^. 
^ 2 2 

Similarly, starting with cos (a + /3) and cos (a — ft) and per- 
forming the same operations, the following formulas result : 



208 MATHEMATICAL ANALYSIS [VII, § 141 

cos P + COS Q = 2 COS ^-±-2 COS ^^-=-2 
2 2 

COS P — COS p = - 2 sin "j" ^ sin ^ » 

2 2 

In words : 

the sum of two sines = 

twice sin (half sum) times cos. (half difference), 
the difference of two sines = 

twice cos (half sum) times sin (half difference),* 
the sum of two cosines = 

twice cos (half sum) times cos (half difference), 
the difference of two cosines = 

minus twice sin (half sum) times sin (half difference).* 

Example 1. Prove that 

cos 3^ +cosa;^^^^g 
sin 3 a; + sin x 

for all angles for which both members are defined. 

cos 3 a; + cosx _ 2 cos ^ (3 x + a:) cos \ (3 a; — a;) _ cos 2 a; _ ^ « ^ 
sin 3 X + sin x 2 sin \ (3 x + a:) cos ^ (3 x — x) ~ sin 2 x ~ 

Example 2. Reduce sin 4 x + cos 2 x to the form of a product. 
We may write this as sin 4 x + sin (90° — 2 x), which is equal to 

2 ,i„ 4a;4-90°-2» ^„3 4»-90° + 2a; ^ ^ ^„ ^^^. + x) COS (3 « - 45°). 

2 2d 

EXERCISES 

Reduce to a product : 

1. sin 4 ^ — sin 2 ^. 4. cos 2 ^ + sin 2 6. 7. cos 3 x + sin 5 x. 

2. cos + cos 3 d. 5. cos 3 ^ — cos 6 6. 8. sin 20° — sin 60°. 

3. cos65 + cos2^. 6. sin (x 4- Ax) — sin X. 

Show that 

9. sin 20° + sin 40° = cos 10°. ^^ sin 15° + sin 75° _ _ ^^ g^o 
10. cos 50° + cos 70° = cos 10°. * sin 16° - sin 75° 

a. sin 75° -sin 15° ^^^3Qo^ 13 sin 3 g- sin5g^_ ^^^^^^ 
cos 75° + cos 15° cos 3 6 — cos 50 

* The difference is taken, first angle minus the second. 



VII, § 142] TRIGONOMETRIC RELATIONS 209 

Prove the following identities : 

14. sin 4 ct -f- sin 3 ct _^^^ce jg sin « + sin /3 __ tan ^ (ce -H3) 

cos 3 a — cos 4 a 2 ", sina — sin/3~tau ^ (a — ^) 

4 e cos a + 2 cos 3 a + cos 5 a cos 3 a 

JLO. ■ ■ = • 

cos 3 « + 2 cos 5 a + cos 7 a cos 5 a 
^rj cosct-cos/3 _ tan|(« + /3) ^g sin (n - 2) g + sin w g _ ^^^ ^ 
cos a + cos ^ ctn i (a -|8) ' cos (n— 2)-0 - cosn^ 

Solve the following equations : 

19. cos 6 4- cos 5 ^ = cos 3 ^. 22. sin 4 — sin 2 ^ = cos 3 d. 

20. sin ^ + sin 5 ^ =: sin 3 d. 23. cos 7 ^ — cos ^ = — sin 4 6. 

21. sin 3 ^ + sin 7 ^ = sin 6 6. 

142. Law of Tangents. A method for shortening computa- 
tion will be presented in the next chapter. In applying this 
method to the solution of triangles the formulas given below 
are valuable. We shall state first the so-called law of tangents: 

The difference of two sides of" a triangle is to their sum as the 
tangent of half the difference of the opposite angles is to the tan- 
gent of half their sum. 

Proof. a^sin^, 

5 sin 5 • 

Hence, by proportion, we have 

g — 6 _ sin ^ — sin ^ 
a 4- 6 sin ^ -f- sin 5 
But 

. ^ . ^ 2cos^i±^sin^:^ tan^^ 
sm ^ - sm J5 2 2 2 



sin ^-f sin ^ o - A -\- B A — B . A 

~ 2 sm — cos tan — 

2 2 

tan^^ 

Therefore a-h^ 2_^ 

« + «> tan4+-§ 



D 



210 MATHEMATICAL ANALYSIS [VII, § 143 

143. Angles of a Triangle in Terms of the Sides. Con- 
^ struct the inscribed circle of the triangle 

and denote its radius by r. If the perim- 
eter a + &4-c=2s, then (Fig. 130) 

AE = AF=s- a, 

.U-s-a^F B BD= BF= s-b. 

Fig. 130. CD = CE = s - c. 

Then tani^ = -^, tani.5 = -^^, taniC=^l-, 
s—a s — b ^ s—c 

where, from § 130, 




-■ J(^-«)(g-^)(^-c) 



' MISCELLANEOUS EXERCISES 

1. Reduce to radians 65°, — 135°, — 300°, 20°. 

2. Reduce to degrees tt, 3 tt, — 2 tt, 4 tt radians. 

3. Find sin (a — /3) and cos (a + ^3) when it is given that a and /3 are 
positive and acute and tan a = | and sec p = ^. 

4. Find tan (a + /3) and tan (a — /3) when it is given that tan a = ^ 
and tan /3 = ^, 

6. Prove that sin 4 a = 4 sin a cos a — 8 sin^ a cos a. 

2 
6. Given sin ^ = — ^, and d in the second quadrant. Find sin 2 d, 
V5 
cos 2 ^, tan 2 ^. 

Prove the following identities : 

. 7. sin2a = -2iHL^. 8. cos2 « = ^ " ^^"'« 

1 -f tan2 a 1 + tan'-^ a 

9. sec2a = _5?^i«_ 10. tan«= "^"^" . 

csc2 a — 2 1 + cos 2 a 

11. sin (a -\- j3) cos /3 — cos (a + p) sin /S = sin a. 

12. sin 2 a + sin 2 )3 + sin 2 7 = 4 sin a sin /S sin 7, if a + j9 + 7 = 180°. 

1 + tan - 

la co8« _ 2 

1 — sin a I .. a 



VII, § 143] TRIGONOMETRIC RELATIONS 211 



14. 1 + tan a tan ^ 



15. 



2 

sin^ a + cos^ ct _ 2 — sin 2 ce 

sin a + cos a " 2 



16. «-H?l^ = 2cos2«. 
sin 2 a 

17. Arc cos f 4- Arc tan f = Arc tan W, 
Solve the following equations : 

18. cos 2 a = cos^ a. 

19. 2 sin a = sin 2 a. 

20. cos 2 a + cos a = — 1. 

21. sin a + sin 2 a + sin 3 a = 0. 

22. sin 2 a — cos 2 a — sin a 4- cos a = 0. 

23. Arc tan x + Arc tan (1 — x) = Arc tan |. 



3 

26. Arc tan ?-tl + Arc tan ^^ = 180° + Arc tan ( - 7). 
x — \ X 

26. Arc sin x + Arc sin- = 120°. 

2 

2 tr 

27. Arcsina; + 2 Arccosx = -^• 

3 

In a right triangle ABC, right angled at (7, prove 

28. sin2 ^ = £jZi? . 29. cos^ ^ = .^±i: . 30. *tan ^ " 



2 2c 2 2c 2 a + 6 

31. Solve for x and ?/ the following equations : 

ic sin a + ^ cos a = sin a, 
X cos a — ?/ sin a = cos a. 

32. Solve for x and y the following equations : 

a; cos ^ — y sin ^ = sin ^, 
X sin ^ + y cos ^ = cos d. 

33. If 2 X is less than 90° and sinx=cos(2 x + 40°), find the value of x. 

34. Find « so that the equation x^ + 2 x cos a + 1 = shall have equal 
roots, 

35. Find a so that the equation 3 x^ + 2 x sec a + 1 = shall have 
equal roots. 



CHAPTER VIII 
THE LOGARITHMIC AND EXPONENTUL FUNCTIONS 

144. The Invention of Logarithms. In the last two chap- 
ters we have had occasion to do a considerable amount of 
numerical computation. In spite of the fact that we have 
confined these computations to comparatively small numbers 
and have had the assistance of tables of squares and square 
roots, the calculations have often been laborious. 

To carry out by the methods thus far at our disposal the 
computations involved in many of the problems of insurance, 
engineering, astronomy, etc., would require a prohibitive 
amount of labor. That it is now practicable jjo effect such 
computations is largely due to the invention ©f logarithms by 
John Napier (1550-1617), Baron of Merchjfeton, in Scotland. 

As in the case of many epoch-making inventions, the funda- 
mental idea of Napier was extraordinarily simple. It may be 
explained as follows. Consider the function y = 2". We 
readily obtain the following table of corresponding values : 



(1) 



X 


1 


2 


3 


4 


6 


6 


7 


8 


9 


10 


11 


12 


2/ = 2^ 


2 


4 


8 


16 


82 


64 


128 


256 


512 


1024 


2048 


4096 



Now, since 2" • 2' = 2"+'', it is clear that, if we desire to ob- 
tain the product of two numbers in the lower line of the table, 
we need only add the two corresponding numbers in the upper 
line (the exponents), and then find the number in the lower 

212 



VIII, § 145] EXPONENTS — LOGARITHMS 213 

line which corresponds to this sum* For example, to find the 
product of 128 x 16, we find from the table that the numbers 
corresponding to 128 and 16 are 7 and 4, respectively; the sum 
of the last pair is 11 and the number in the lower line corre- 
sponding to 11 is 2048, which is the product sought. Or again, 
to find 4096 -?- 512, we find the corresponding exponents 12 and 
9 in the table, subtract (12 — 9 = 3), and find the required quo- 
tient to be 8. How would you justify the latter procedure ? 

While the fundamental idea here described is simple, con- 
siderable insight was required to make the idea practicable. 
For, the above table makes possible the finding of the product 
of two numbers only when the numbers in question and 
their product are to be found in the lower line of the table. In 
order to be useful in practical computation it is obviously 
necessary to construct a table which will contain every number, 
or at least from which the corresponding " exponent " of any 
number can easily be obtained either precisely or with a high 
degree of approximation. Th^ problem confronting Napier 
was to Jill in the gaps in the numbers of the lower line of the 
table on p. 212, while preserving the fundamental property of 
the table, yIz. that to the product of any two numbers of the lower 
line corresponds the sum of the two corresponding numbers of the 
upper line. 

145. Extension of the Table. An examination of table (1) 
reveals the following properties : (a) the values of x form an 
arithmetic progression (A.P.), since every number after the 
first is obtained by adding 1 to the preceding number ; (6) the 
values of y form a geometric progression (G.P.), since every 
number after the first is obtained by multiplying the preceding 
number by 2. These considerations suggest the possibility of 
extending the table in two ways. 



214 



MATHEMATICAL ANALYSIS [VIII, § 145 



-5 


-4 


-3 


-2 


- 1 

• 





1 


2 


o 


4 


5 


6 


1 

7 


0.03126 


0.0625 


0.125 


0.25 


0.5 


1 


2 


4 


8 


16 


32 


64 


128 



In the first place, we may extend it to the left so as to make 
the lower line contain numbers less than 2. To do this, we 
need only subtract 1 successively from the numbers of the upper 
line and divide by 2 successively the numbers of the lower 
line. We then obtain a table extending in both directions : 



(2) 



This table is still satisfactory. If we desire to multiply 128 
by 0.0625, we add the corresponding numbers of the upper line, 
namely, 7 and — 4 ; thus we obtain the number 3, which 
according to the previous rule should give 128 x 0.0G25 = 8, 
which is correct. That the rule still applies may be tested on 
other products ; the fact that it does will be proved later. 

In the second place we may find new numbers to fill the gaps 
in the original table, by inserting arithmetic means between 
the successive values of x and geometric means between the 
successive values of y. Thus, if we take the following portion 
of the preceding table 



-2 


-1 





1 


2 


3 


4 


i 


i 


1 


2 1 4 


8 


IG 



and insert between every two successive numbers of the upper 
line their arithmetic, and between every two successive num- 
bers of the lower line their geometric mean, we obtain the table 



(3) 



-2 


-f 


- 1 


-h 





i 


1 


3 
2 




5 

-2 


3 


1 


4 


i 


^\/2 


h 


iV2 


1 


V2 


2 


2V2 


4 


4V2 


8 


8V2 


16 



VIII, § 145] EXPONENTS -— LOGARITHMS 



215 



If the radicals are expressed approximately as decimals, this 
table takes tlie form 



-2.0 


-1.5 


- 1.0 


-0.5 


' 


0.5 


1.0 


1.5 


2 


2.5 


3 


3.5 


4 


0.25 


0.35 


0.50 


0.72 


l.OC 


1.41 


2.00 


2.8S 


4.00 


5.66 


8.00 


11.31 


16 



x(AP.) 


0.00 


0.25 


0.50 0.75 


1.00 


1.25 1.50 


1.75 


2.00 


2.25 


yCG.R) 


1.00 


1.19 


1.41 ' 1.68 


2.00 


2.38 2.83 


3.36 


4.00 ' 4.76 

1 



Repeating this process of inserting means, we get the follow- 
ing table. To save space, we have begun the arithmetic pro- 
gression with and the geometric progression with 1, and have 
not carried the table as far as in the preceding case. 



(4) 



The rule for multiplying two values of y seems to apply also 
to this table, at least approximately. For example, if we apply 
the rule to find 3.36 x 1.19, we note that the sum of the cor- 
responding values of x is 1.75 -f 0.25 = 2.00 and conclude that 
3.36 X 1.19 = 4.00. Actual multiplication gives 3.36 x 1.19 
= 3.9984. The discrepancy we may attribute to the fact that 
the values of y other than 1, 2, 4 are only approximations to 
the true values.* 

The process used in constructing this table may be continued 
indefinitely. It enables us to interpolate a new value of x be- 
tween any two successive values of x and a new value of y 
between the two corresponding values of y. But this means 
that we can make the values of x and y as dense as we please, 
in other words, we can make the difference between successive 
values of y as small as we please. By continuing the process 

♦ In fact the rules for computing with approximate numbers would lead us 
to write 4.00 in place of 3.9984 as we have no right to retain more than two 
decimal places. See § 160. 



216 



MATHEMATICAL ANALYSIS [VIII, § 145 



long enough we can make any number appear among the values 
of y to as high a degree of approximation as we desire and our 
intention of filling the gaps will then be attained. We must 
now prove, however, that the rule for multiplication does really 
hold in the extended table. Thus far we have merely verified 
this rule for special cases. 

EXERCISES 

1. Assuming that the rule for multiplication applies, find by means of 
table (4) the following products. 

3.36 X 1.41, 1.68 X 2.38, (1.68)2, (1.19)6. 
Check by ordinary multiplication. 

146. Arithmetic and Geometric Progressions. The tables 
constructed consist of an arithmetic progression one term of 
which is the number (the terms of this arithmetic progression 
we denoted by x) and a geometric progression one term of 
which is the number 1 (the terms of this geometric progression 
we denoted by y). Moreover, to every value of x corresponds a 
definite value of y in such a way that to oj = corresponds 2/ = 1, 
and that to each succeeding (or preceding) value of x corre- 
sponds the succeeding (or preceding) value of y. Now suppose 
that the common difference of the arithmetic progression is d 
and that the common ratio of the geometric progression is r.- 
The correspondence between the values of x and y would then 
be exhibited in the following table. 



(5) 



We shall now prove that in this tahle^ to the product of any 
two valiLes of y corresponds the sum of the two corresponding 
values ofx. 



X 


... 


-md 


... 


-3d 


-2d 


-d 





d 


2d 


Sd 


... 


nd 


... 


y 


... 


1 


... 


1 


1 


1 
r 


1 


r 


r^ 


r8 




r" 


... 



Vni, § 147] EXPONENTS — LOGARITHMS 



217 



If tlie two values of y are both, to the right of y = 1, for ex- 
ample i/i = r^, y.2 = J*^, then the corresponding values of x are 
pd and qd. To the product 2/12/2= ^^^'^ corresponds {p-\-q)d. 
If the two values of y are both to the left of 2/ = 1, the proof is 
similar. It is left as an exercise. 

If one value is to the left of 2/ = 1, for example, y — l/r^, and 
the other value is to the right, for example 2/2 = ''^j the cor- 
responding values of x are — pd and qd respectively. The 
product 2/12/2 is equal to (1/r^) r« = r^~^ if q> p, and is equal 
to l/r^« if q<p. The value of x corresponding to 2/12/2 is 
then {q — p)d/\i q> p and —{p — q)d ii q<,p. But {q — p)d 
= — {p —q)d = qd -\-{— pd). The discussion of the case 
p = q\^ left as an exercise. If one of the values of y is 1, the 
desired result follows immediately. Why ? 

In view of this theorem the validity of the rule used in the 
last article for multiplication is established. For tables (2) 
and (4) are both tables of the type (5), the former having 
d = 1 and r = 2, the latter having d — 0.25 and r = V2 = 1.19 
(approximately) . 

147. The Exponential Function (i^{a > 0). Let us now con- 
sider the table 



X 


— m 


... 


-3 


-2 


- 1 





1 


2 


3 


... 


n 


... 


y 




... 


1 
a3 


1 
a2 


1 
a 


1 


a 


a2 


a^ 




a'» 


... 



where a represents any positive number.* This table defines y 
as a function of x. Morover, this table is a table of the type 
(5) ; and all tables obtained by interpolating arithmetic means 
between two successive values of x and the same number of 
geometric means between the corresponding values of y are of 

* The value a = 1 leads to trivial results. Hence, we assume also that a ^ 1. 



218 



MATHEMATICAL ANALYSIS [VIII, § 147 



the type (5). Thus if we interpolate q arithmetic means be- 
tween x — and x = 1, and q geometric means between y —1 
and y= a, we obtain the following table : 



X 


... 




... 


-1 




-2 


_1 





1 
Q 


2 1 ... 


y 


... 


1 


... 


1 
a 




1 


1 


1 


la 


^.'/;;^2 ... 


V^f 


i 1 


X 




1 


9 + 1 
Q 


... 


2 






... 


. 


iv'ar' 


a 


{Vay^' 


... 


a2 




{VaY 


... 



which is a table of the type (5), with d = 1/q and r = va. 

The function y oi x thus defined is y = a% for ic = 1, 2, 3, •••. 
We are therefore led to define the expression a* for fractional 
and negative values of x and for x = as follows : 

(1) aO = l. 

(2) ai/« means ^o", where q^ is a positive integer. 

(3) aP/^ means (Va)^, or its equal V^y where p and q 
are positive integers. 

(4) a-^ means 1/a**, where n is any positive rational number. 
In view of the fundamental property of any table of type 

(5), whereby to the product of any two values of y corresponds 
the sum of the two corresponding values of x, we have 

(t^ . av — flM+f 
for^all values of u and v for which the expressions a", a", and 
a""*"' have been defined. 

The function ?/ = a* (a > 0) has now been defined for all 
rational values of x. To complete the definition of this func- 

* We should keep in mind that the symbol ^a (o>0) means the positive 
gth root of a. Thus yfm = 2, not — 2. 



VIII, §147] EXPONENTS — LOGARITHMS 219 

tion for all real values of a;, we must indicate the mean- 
ing of a* when x is an irrational number. To carry this 
definition through in all its details is beyond the scope of 
an elementary course. But we have seen that any irrational 
number may be represented approximately by a rational num- 
ber, with an error as small as we please. (See § 29.) Thus 
V3 is represented approximately by the rational numbers 1.7, 
1.73, 1.732, ..-. Our previous definitions have given a definite 
meaning, for example, to 2^-^, 2^-''^, 2'^-''^^, .... The values of the 
latter expressions are by definition approximate values of 2^^. 
We take for granted without proof the fact that the successive 
numbers 
(6) 2^'\ 21-7^ 21-"^, ..., 

as the exponents represent closer and closer approxima- 
tions to V3, approach closer and closer to a definite number. 
This definite number is by definition the value of 2^. Similar 
considerations apply to the definition of a^, where a is any 
positive number and x is any irrational number. The principle 
involved is briefly expressed as follows : 

An approximate value of x gives an approximate value of a'. 
The value of a* can he found as accurately as we please by using 
a sufficiently accurate approximation to x. 

The objection might be raised that the calculation of '2?''^ involves the 
extraction of the 10th root of 2 and the calculation of 2^-'^ involves the 
extraction of the 100th root of 2, etc., and perhaps we do not know how 
to extract these roots. As a matter of fact we can calculate 2^^ as ac- 
curately as we please by extracting square roots only. The processus as 
follows : We know that \/3 = 1.7320 accurately to four decimal places. 
Now by table (4), p. 215, we see that 2^-^-2M and 2'-" = 3.36. We carry 
the computation to more places and have 2i«)oo =2.8284 and 2i-7500= 3.3636. 

Now, 1.7320 lies between 1.5000 and 1.7500, the arithmetic mean of 
which is 1.6250. The geometric mean of 2.8384 and 3 3635 is 3.0844 
According to our previous definitions we have then 2i-<'25o = 3.0844. 



220 MATHEMATICAL ANALYSIS [VIH, § 147 

Inserting means between the last two results we have 2^-^^ = 3.2200. 
By inserting arithmetic means between the properly selected exponents 
and geometric means between the corresponding powers of 2 we can ulti- 
mately obtain the value of 2i-^320, ^he results of the necessary steps are : 
21.7188 :^ 3.2915, 21-7344 = 3.3274, 21-7266 = 3.3094, 2i-7305 = 3.3182, 
21.7325 == 3.3228, 21-7315 = 3.3205, 2i-7320 = 3. 32 17. 
The process here illustrated makes it possible to calculate 2^3 to as high 
a degree of approximation as we please, since we can carry the computa- 
tion to as large a number of decimal places as we please. 

148. The Laws of Exponents. The function y = a'' (a > 0) 
is now defined for all real values of x. This function is called 
the exponential function of base a. The laws of exponents 

I. a^ ' a^ = a^^^ ] 

II. (cr*)'' = a«*» , a > 0, 6 > 0, 

III. a«* . b^ = (ab)^ ' 

which were derived previously (§ 42) for positive integral ex- 
ponents, hold for all real values of u and v. The first of these 
we have already derived. The last two may be readily proved 
for negative, fractional, and zero exponents by using the defini- 
tion of a''. 

Thus by definition, if it = p/q and v = 71, where p, q, n are 
positive integers, we have 

p pn 

If u is any positive rational number and v=p/qf where p, q are 
positive integers, we have, 

up 



(a^y = (a«)« = -Via'^y = -^/a"^ = a «" = a"". 
If u = — n, where n is a positive rational number, and if 
V =p/q, where p and q are positive integers, we have 

(a-y = (a-«)^/'=/^\/iY= — i— = -i = a" 7 = a«^ 
If u is any rational number and v =— n, where n is any 



VIII, § 149] EXPONENTS — LOGARITHMS 
positive rational number, 



221 



(a«)''=(a'*)-" 



— — — = — = a""" = a* 



If either w or v is zero, the result is immediate. Hence the law 
II is proved for rational exponents. 

A similar proof of the law III is left as an exercise. 

149. The Graph of the Exponential Function. Figure 131 
represents the graph of the function y = 2*, drawn from the 
tabular representation given in the first table on p. 215. 







~ 


- 


7 




















r 








^ 
















~ 












n 






































































\ 




















l. 






1 




























\ 












u 










































\ 






[e 












ri//^/^ 




h 












-12 
















\ 






R 


- 










«l \\ 


.U 




's 
























i 






















\ 




i 




























/ 
























\ 




/ 




























/ 














' 












/ 




























/ 






?^ 






















/ 


























y 


























j 




























/ 








"^ 














/ 




1 


























/ 


r 












\ 












' 


/ 








/ 




















/ 














\\ 








/ 






/ 






/ 




















/ 










<^ 


\ A 


\ 










/ 




y 


<\ 




















/ 


' 1 








• 




'5' 


N 


s^ 






V. 


/ 




X 


"*" 
















1 




y 






















^ 


^ 




^ 








a 


= 1 








^ 
























1 








^ 


^ 


?J 


, 


_ 


—J 








-4 












































' 
























V 


=2tc 














































































V 


= 


^ 


_ 











Fig. 131 



FiQ. 132 



It will be noted that all curves of the system ?/ = a* pass 
through the point (0, 1). By hypothesis a > 0. If a > 1, the 
function a^ is an increasing function ; while if a < 1, the func- 
tion is a decreasing function. Figure 132 shows some of the 
curves of the system y = a*. 



222 MATHEMATICAL ANALYSIS [VIII, § 149 

EXERCISES 

1. Calculate the value of the exponential function 3* 

(a) for the values of a; = 1, 2, 3, 4, 0, — 1, — 2, - 3, — 4 ; 

(b) for the values x = 0.5, 1.5, 2.5, 3.5, - 0.5, - 1.5, - 2.5 ; 

(c) for the values x = 0.25, 0.75, 1.25, 1.75. 
Arrange the results in the form of a single table. 

2. Show how to use the table constructed in Ex. 1 to solve problems in 
multiplication, division, raising to powers, and extracting roots. Make up 
your own problems and check your results by the methods of arithmetic. 

3. Describe in detail how you would find the value of 3 . Between 
what two numbers in the table found in Ex. 1 does the value of 3^^ lie ? 

4. Construct the graph of y = 3* for values of x between — 2 and 3. 
6. What is meant by a^ ? x^ ? (l/y)^ ? 

6. What is the value of 8^ ? 27^ ? (0.001)^ ? (i)8 ? 

7. Simplify (18)^ -i- (3)^' 

8. Perform the following indicated operations : 

(«) (^¥. w (^>^-K 



\x!^y~^y 
(6)(aWc¥. (,)^^_,+ 2,.i).. 

(c) (32x02/10)^. (/) (2^)*. 

9. Multiply 
(a) {a-^ + a)ia-^-a). (6) (a-i-ao)(a-2-a'0(a-»-aO- 

(c) (a;^-y^)(a;« -y^). 

(d) (x-i 4- x~^y~^ + jr^Xx-i - x~^y~^ + y-^). 

8 2 12 1 

(e) (a* - 2 a* + 3 a*) (2 a'- a* + 2). 

8 2 12 1 

(/) (ir - oir + 3 6y« - c) (jr + 6y« - cyO). 

10. Divide 

(a) (a;+l)by (v^ + 1). (&) (aj^' - y*) by (x^- yi^). 

(c) (J - a6^ + ah - b^) by (a^ - 6^). 

(d) (a-i + 4 a^ + 6 a^ + 4 a^ + a*) by (a"^ + a). 



VIII, § 150] EXPONENTS — LOGARITHMS 223 

11. Simplify 
(a) 12" + I - 9-' + ^ + 27*. (6) ( '"'P ]'*. 



^64 m-^p^ 
12. Simplify 



13. Which of the two numbers V5 and ^/8 is the greater and why ? 

14. Simplify 

(2^ x2^)-f-(54)i 

15. Prove that, if 



_ 1 Tx^ _ x"^"| 

2Lj „-d' 



■y y 

then 

2vxy 
16. Reduce to simplest form 

(c) (a^ + X*) (a2 _ x^)~^ - (a2 _ x^)^. 

150. Definition of the Logarithm. The logarithm of a 
number JV to a base 6 (6 > 0, ^^t: 1) is the exponent x of 
the power to which the base h must be raised to produce the 
number N. 

That is, if 

then 

X — logbN. 

These two equations are of the highest importance in all work 
concerning logarithms. One should keep in mind the fact 
that if either of them is given, the other may always be 
inferred. 



224 



MATHEMATICAL ANALYSIS [VIII § 150 



Tlie graph of the logarithm function (Fig. 133) is obtained 
from the graph of the corresponding exponential function by 
simply turning the latter graph over about the line through 
the origin bisecting the first and third quadrants. 









Y 




































- 


























































, 












































































<- 


.— ■ 


* 












































































^ 


^ 










































^ 


x" 




































'" 






















































/ 










































^ 




/ 
















































f 

















































1 J 
















1 


> 








1 


s 








^^x\ 






■ 1 
























































V 


= z 


ff^ 


X 






















































































. 








































_ 














_ 








_J 




















_ 







Fig. 133 



EXERCISES 

1. When 3 is the base what are the logarithms of 9, 27, 3, 1, 81, ^, 

2. Why cannot 1 be used as the base of a system of logarithms ? 

3. When 10 is the base what are the logarithms of 1, 10, 100, 1000 ? 

4. Find the values of x which will satisfy each of the following 
equalities : 

(a) logs 27 = X. (d) loga a = x. (g) logs x = 6. 

(6) log^3 = 1. (e) logal = x. {h) logssx = i 

(c) log, 6 = i. (/) logaij^r = «• (0 log.ooi x = .00001. 

6, Find the value of each of the following expressions : 

(a) loga 16. (c) logesi-g. (e) log26l26. 

(6) log34a49. (d)log2Vl6. (/) log2^. 



VIII, § 151] EXPONENTS — LOGARITHMS 225 

151. The Three Fundamental Laws of Logarithms. From 
the properties of tlie exponential function (p. 220) we derive 
the following fundamental laws. 

I. Tlie logarithm of a product equals the sum of the logarithms 
of its factors. Symbolically, 

lege, MN = logft Af + logft N. 

Proof. Let logj, M= x, then 6* = M. Let log^ N~y, then 
b"=N. Hence we have MN= ft*"^", or 

logj,i¥iV=a; + 2/, i.e. \o^^,MN=\og^,M+\o^^N. 

II. The logarithm of a quotient equals the logarithm of the 
dividend minus the logarithm of the divisor. Symbolically 

M 
^^Sft TT = logft M — logft N. 
N 

Proof. L^t logj, Jf = x, then b' = M. Let logj, -^ = y, then 
h" = N. Hence we have M/N = &*"»', or 

M M 

^og,—=x-y, i.e. log, — = log,Jf-log,JV 
IIL T%e logarithm of the pth power of a number equals p 
times the logarithm of the number. Symbolically 

log6MP = />log5M. 

Proof. Let log^ M^x, then 6^ = M. Raising both sides 
to the pih. power, we have b^ = M^. Therefore 

logj, M^ =px=p log, M. 

From law III it follows that the logarithm of the real positive 
nth root of a number is one nth of the logarithm of the number. 

Q 



226 MATHEMATICAL ANALYSIS [VIII, § 151 

EXERCISES 

1. Given logio 2 = 0.3010, logio 3 = 0.4771, logio 7 = 0.8451, find the 
value of each of the following expressions: 

(a) logio 6. (/) logio 6. 

[Hint: logio 2 X 3=logio 2+logio3.] [Hint : logio5 = logio V-] 

(6) logio 21.0. (.9) logio 150. 

(c) logio 20.0. W logio Vli. 

{d) logio 0.03. (i) logio 49. 

(«) logio 1. U) logio V2V7^. 

2. Given the same three logarithms as in Ex. 1, find the value of each 
of the following expressions: 

f^\ i«„ 4 x5 x7 ., . i^„ 5 X 3 X 20 ,. , ^ 2058 

{d) logio (2)25. (e) logio (3)8(5)6. (/) logio (28)(|). 

152. The Systems most Frequently Used. From the defi- 
nition of a logarithm (§ 150) any positive number except 1 can 
be used as the base of a system of logarithms. As a matter of 
fact, however, the numbers generally used are (1) a certain 
irrational number which is approximately equal to 2.71828 
and is denoted by e and (2) the number 10. Logarithms to the 
base 6 are important in certain theoretical problems ; loga- 
rithms to this base are called natural. For numerical compu- 
tation it will be seen presently that the base 10 has numerous 
advantages. Since different systems of logarithms are in use, 
it is important to know how to change from one system to 
another. The following law explains how this can be done. 

IV. The logarithm of a number M to the base b is equal to the 
logarithm of M to any base a, divided by the logarithm of b to the 
ba^e a. Symbolically, 

lOga O 



VIII, § 153] EXPONENTS — LOGARITHMS 227 

Proof. Let logj, M=x, then ¥ — M. Taking the logarithms 
of both sides to the base a, we have 

log^ ?>=^ = log„ 3f, or a; log„ 6 = log, 3f, 

log« h 

153. Logarithms to the Base 10. Logarithms to the base 
10 are known as common logarithms, or as Briggian logarithms, 
after Henry Briggs (1556-1631) who called attention to the 
advantages of 10 as a base. These advantages appear below. 

If 10 is the base, log 10 = 1, log 100 = 2, log 1000 = 3, etc. 
It follows that if a number be multiplied by 10, or by any 
positive integral power of 10, the logarithm of the number is 
increased by an integer. In other words, the shifting to the 
right of the decimal point in a number changes only the in- 
tegral part of the logarithm and leaves unchanged the decimal 
part of the logarithm. 

An example will make this clear. Given logio 2 = 0.30103, we have 
logio 20 = 1.30103, logio200 = 2.30103, etc. Or, again, given logio 4.5607 
= 0.65903, we have logio 45.607 = 1.65903, logio 456.07 = 2.65903, 
logio 4560.7 = 3.65903, logio 45607.0 = 4.65903. 

It should be clear from these examples that the decimal part 
of the logarithm of a number greater than 1 in this system 
depends only on the succession of figures composing the num- 
ber, irrespective of where the decimal point is located ; while 
the integral part of the logarithm of the number depends 
simply on the position of the decimal point. 

The decimal part of a logarithm is called its mantissOj the 
integral part its characteristic. In view of what has been said 
above only the mantissas of logarithms to the base 10 need be 
tabulated. The characteristic can be found by inspection. 
This follows from the following considerations. 



228 MATHEMATICAL ANALYSIS [VIII, § 153 

The common logarithm of a number between 1 and 10 lies 
between and 1. 

The common logarithm of a number between 10 and 100 lies 
between 1 and 2. 

The common logarithm of a number between 100 and 1000 
lies between 2 and 3. 

The common logarithm of a number between 10" and lO'*'^^ 
lies between n and n + 1. 

It follows that a number with one digit (^0) at the left of 
the decimal point has for its logarithm a number equal to + a 
decimal ; a number with two digits at the left of its decimal 
point has for its logarithm a number equal to 1 + a decimal ; a 
number with three digits at the left of the decimal point has 
for its logarithm a number equal to2-f- a decimal, etc. We 
conclude, therefore, that the characteristic of the common loga- 
rithm of a number greater than 1 is one less than the number of 
digits at the left of the decimal point. 

Thus, as before, logio 456.07 = 2.65903. 

The case of a logarithm of a number less than 1 requires 
special consideration. Taking the numerical example first con- 
sidered above, if logjo 2 =0.30103, we have logio 0.2 = 0.30103-1. 
Why ? This is a negative number, as it should be (since the 
logarithms of numbers less than 1 are all negative, if the 
base is greater than 1). But, if we were to carry out this 
subtraction and write logjo 0.2 = — .69897 (which would be 
correct)^ it would change the mantissa, which is inconvenient. 
Hence it is customary to write such a logarithm in the form 
9.30103 - 10. 

If there are n ciphers immediately following the decimal 
point in a number less than 1, the characteristic is — n — 1. 
For convenience J if n < 10, we write this as (9 — n) — 10. This 



Vni, § 154] EXPONENTS — LOGARITHMS 229 

characteristic is written in two parts. The first part 9 — n is 
written at the left of the mantissa and the — 10 at the right. 

In tlie sequel, unless the contrary is specifically stated we 
shall assume that all logarithms are to the base 10. We may 
accordingly omit writing the base in the symbol log when there 
is no danger of confusion. Thus, the equation log 2 = 0.30103 
means logjo 2 = 0.30103. 

154. Use of Tables. Since the characteristic of the loga- 
rithm of a number may be found by inspection, a table of 
logarithms contains only the mantissas. To make practical 
use of logarithms in computation it is necessary to have a con- 
veniently arranged table from which we can find (a) the 
logarithm of any given number, and (6) the number corre- 
sponding to a given logarithm. Tables of logarithms differ 
according to the number of decimal places to which the man- 
tissas are given and also in incidental details. However, the 
general principles governing their use are the same. These 
principles are explained for a four-place table (p. 536) by the 
following examples. 

Problem A. To find the logarithm of a given number, 

(1) When the number contains three or fewer figures. 

Example. To find the logarithm of 42.7. 

First, by §153, the characteristic is 1. We accordingly write (provi- 

^'°°^'y^ log42.7 = l. 

Next we look up in the tables the mantissa corresponding to the succes- 
sion of figures 4, 2, 7. We run a finger down the first column of the 
table until we reach the figures 4, 2, hold it there while with another 
finger we mark the column headed with the third figure, 7. At the 
intersection of the line and column thus marked, we find the desired 
mantissa : 6304, The desired result is then 

log 42.7 = 1.6304. 



230 MATHEMATICAL ANALYSIS [VIII, § 154 

To find the logarithm of 0.0427, we should proceed in precisely the 
same way, the only difference being that the characteristic is now 8 — 10. 

H®^<^®' log 0.0427 = 8.6304 - 10. 

(2) When the number contains four significant figures 

Example. To find log 32.73. 

We see that again the characteristic is 1, and we write provisionally 
log32.73 = l. 
Now, the mantissa of log 32.73 lies between the mantissas of log 32. 70 and 
log 32.80; i.e. (from the table) between 5145 and 5159. The difference 
between these two mantissas (called the tabular difference at that place in 
the table) is 14, und this difference corresponds to a difference in the 
numbers of .10. According to the principle of linear interpolation,* the 
difference in the mantissas corresponding to a difference in the numbers 
of .03 is 14 X .3 = 4.2 or (rounded) 4. The mantissa corresponding to 
3273 is then 5145 + 4 = 6149, and we obtain 
log 32.73 = 1.5149. 

Problem B. To find the number corresponding to a given 
logarithm. Here we simply reverse the preceding process. 

Example. To find the number whose logarithm is 0.8485. 

We first seek the mantissa 8485 in the table. We find that it lies be- 
tween 8482 and 8488, corresponding respectively to the successions of 
figures 7050 and 7060. The tabular difference here is 6, while our differ- 
ence, i.e. the difference we have to account for (8485 — 8482) is 3. 
Hence the corresponding difference in the numbers is | of 10 or 6. Hence 
the succession of figures in the number sought is 7055. Since the char- 
acteristic is 0, the number sought is 7.055. Or, log 7.055 = 0.8485. 

If the mantissa is found exactly in the table, of course no interpolation 
is necessary. Thus the number whose logarithm is 9.7348 — 10 is 0.5430. 

EXERCISES 

1. Find the logarithms of the following numbers from the table on 
pp. 536-7 : 482, 26.4, 6.857, 9001, 0.5932, 0.08628, 0.00038. 

2. Find the numbers corresponding to the following logarithms : 
2.7935, 0.3502, 7.9699 - 10, 9.5300- 10, 3.6698, 1.0958. 

* One should convince oneself that the conditions for linear interpolatiou 
are satisfied by this table. In fact, it is readily seen that for several numbers 
immediately preceding and following 327, the tabular differences are 13 and 14. 



VIII, § 155] EXPONENTS — LOGARITHMS 231 

155. Use of Logarithms in Computation. The way in which 
logarithms may be used in computation will be sufficiently ex- 
plained in the following examples. A few devices often neces- 
sary or at least desirable will be introduced. The latter are 
usually self-explanatory. Reference is made to them here, in 
order that one may be sure to note them when they arise. 
The use of logarithms in computation depends, of course, on 
the fundamental properties derived in § 151. 

Example 1. Find the value of 73.26 x 8.914 x 0.9214. 

We find the logarithms of the factors, add them, and then find the 
number corresponding to this logarithm. The work may be arranged as 
follows : 



Numbers 




Logarithms 


73.26 


(->) 


1.8649 


8.914 


(-» 


0.9501 


0.9214 


(-» 


9.9645 - 10 




12.7795-10 


601.9 ^ws. ' 


(^) 


2.7795 



Product 

Example 2. Find the value of 732.6 ^ 89.14. 

Numbers Logarithms 

732.6 (^^) 2.8649 

89.14 (->) 1.9501 

Quotient = 8.219 Arts. (<-) 0.9148 

Example 3. Find the value of 89.14 -- 732.6. 

Numbers Logarithms 

89.14 (->>) 11.9601 - 10 

732.6 (-^) 2.8649 

Quotient = 0.1217 Ans, (<~) 9.0852 - 10 

Example 4. Find the value of 763.2 x 21.63 

Whenever an example involves several different operations on the 
logarithms as in this case, it is desirable to make out a blank form. When 
a blank form is used, all logarithms should be looked up first and entered 
in their proper places. After this has been done, the necessary opera- 
tions (addition, subtraction, etc.) are performed. Such a procedure 
saves time and minimizes the chance of error. 



232 



MATHEMATICAL ANALYSIS [VIII, § 155 



Form 



Numbers 




Logarithms 


763.2 




(-» 


21.63 




(-» (+)..... 


product 




986.7 


Ans. 


(-» (-) ." '.'.'. 


.... 


(<-) 




Form Filled In 


Numbers 




Logarithms 


763.2 




(->) 2.8826 


21.63 




(->) 1.3351 


product 




4.2177 


986.7 




(->) 2.9942 


16.73 


Ans. 


i<-) 1.2235 


Example 5. Find (1.357)6. 




Numbers 




Logarithms 


1.357 




(->) 0.1326 


(1.357)6 : 


= 4.602 


Ans. (-^) 0.6630 


Example 6. Find the cube root of 30.11. 


Numbers 




Logarithms 


30.11 




(_>) 1.4787 


y/SOAl = 


3.111 


Ans. «-) 0.4929 



Example 7. Find the cube root of 0.08244. 

Numbers Logarithms 

0.08244 (->) 28.9161-30 

v^O.08244 = 0.4352 Ans. (-^) 9.6387 - 10 



EXERCISES 

Compute the value of each of the following expressions using the table 
on pp. 536-537. 

1. 34.96 X 4.65. 5. (34.16 x .238)2. 

2. 518.7 X 9.02 x .0472. 6. 8.572 x 1.973 x (.8723)8. 
„ 0.5683 



0.3216 

6.007 X 2.483 
6.524 X LUO' 



7. 



648.8 



^(21.4)2 
\2791 



Vm, § 155] EXPONENTS — LOGARITHMS 233 

9. 



10. 
11. 



J 2.8076X 3.184 
^ (2.012)3 

»/ 2941 X 17^32 
'>'2173 X 18.76* 

\/0. 00732 




V735 ^^ 

12, (20.027)i. (2.01)i 

17. The stretch s of a brass wire when a weight m is hung at its free 
end is given by the formula , 

where m is the weight applied in grams, g = 980, I is the length of the 
wire in centimeters, r is the radius of the wire in centimeters, and yfc is a 
constant. If m = 844.9 grams, I — 200.9 centimeters, r = 0.30 centi- 
meters when s = 0.056, find k. 

18. The crushing weight P in pounds of a wrought iron column is given 
by the formula -^ 55 

P = 299,600^^, 

where d is the diameter in inches and I is the length in feet. What weight 
will crush a wrought iron column 10 feet long and 2.7 inches in diameter ? 

19. The number n of vibrations per second made by a stretched string 
is given by the relation . , — 

2lM m' 
where I is the length of the string in centimeters, ilf is the weight in grams 
that stretches the string, m the weight in grams of one centimeter of the 
string, and g =980. Find n when i{f= 5467.9 grams, Z = 78.5 centi- 
meters, m = 0.0065 grams. 

20. The time t of oscillation of a pendulum of length I centimeters is 

given by the formula 

t = rJ-L. 
>'980 

Find the time of oscillation of a pendulum 73.27 centimeters in length. 

21. The weight w in grams of a cubic meter of aqueous vapor saturated 
at 17° C. is given by the formula 

^ ^ 1293 X 12.7 X 5 

computer. (1+M)(760x8)- 



234 MATHEMATICAL ANALYSIS [VIII, § 156 

156. Exponential Equations. An equation in which the 
unknown is contained in an exponent is known as an exponen- 
tial equation. Some such equations may be solved by taking 
the logarithms of both sides after the equation has been 
properly transformed. 

2x4-1 

Example 1. Solve the equation 3 + 7 = 15. 

Transposing the 7 and taking logarithms of both sides we obtain 

{2x + l)log3 = log8. 

Hence we find 



^^iH-^^ij. 



2Llog3 



Example 2. Money is placed at interest, compounded annually. Find 
a formula for the amount at the end of n years. Also a formula* giving 
the number of years necessary to produce a given amount. 

Let C be the original capital and r the given rate of interest {i.e. if 

the interest is 5 per cent, r = 0.05). The amount A\ at the end of the 

first year is ^ ^ ^ ^,, 

Ai= G+ Cr= C{\ + r). 

At the end of two years we have 

^2 = ^1(1 + = 0(1 + 02. 

At the end of n years, the amount is 

A=A,= G{\^rY. 

This is the formula required. To find w, given A, O, r, we take the 
logarithms of both sides and find 

EXERCISES 

1. Solve for x the equation 2^ = 5. 

2. Solve for y the equation 3w + 2 = 9. 

■ 3. Solve for x and y the simultaneous equations S^+v = 4, 2*~v = 3. 
4. Solve for x and y the simultaneous equations 2*+f = 6y, 3*-^ = 2^+^ 
6. Find the amount of $1000 in 25 years at 6 per cent compound 
interest, compounded annually. 



VIII, § 1 56j EXPONENTS — LOGARITHMS - 235 

^ 6. Find the amount of $ 600 in 10 years at 4 per cent compound inter- 
est, compounded semiannually. 

7. In how many years will a sum of money double itself at 5 per cent 
interest compounded annually ? semiannually? 

8, A thermometer bulb at a temperature of 20° C. is exposed to the air 
for 15 seconds, in which time the temperature drops. 4 degrees. If the 
law of cooling is given by the formula 6 = ^oe~^S where is the final tem- 
perature, do the initial temperature, e the natural base of logarithms, and 
t the time in seconds, find the value of b. 



MISCELLANEOUS EXERCISES 

1. "What objections are there to the use of a negative number as the 
base of a system of logaiithms ? 

2. Show that a^^g^"^ = x. 

3. Write each of the following expressions as a single term : 

(a) log x-\-\ogy — log z. (6) 3 log a; — 2 log y + 3 log z. 

(c) 3 log a - log (x + y)- I log (ex + d)+ log VwT~x. 

4. Solve for x the following equations : 

(a) 2 log2 X -h log2 4 = 1. (c) 2 logio x — 3 logio 2 = 4. 

(b) logs « - 3 logs 2=4. {d) 3 logs x + 2 loga 3=1. 

5. How many digits are there in 235 ? 3142 ? 312 ^28? 

6. Which is the greater, (il)^^^ or 100 ? 

7. Find the value of each of the following expressions. (See § 152. ) 
(a) logeSS. (6) logs 34. (c) log7 246. (cZ) logi3 26. 

8. Prove that logb a • logo b = 1. 

9. Prove that 

log, x-{-Vx^-l ^ 2 log, lx+ v^^^l]. 

X — y/x'^ — 1 

10. The velocity v in feet per second of a body that has fallen s feet is 
given by the formula v = \/64.3s. 

What is the velocity acquired by the body if it falls 45 ft. 7 in. ? 

11. Solve for x and y the equations : 2^ = IGv, aj + 4 y = 4. 



CHAPTER IX 

NUMERICAL COMPUTATION 

I. ERRORS IN COMPUTATION 

157. Absolute and Relative Errors. In § 29 we noted that 
the numerical result of every observed measurement is an 
approximation. The difference between the exact value of 
the magnitude and this observed value is a concrete number 
called the absolute error* Often the absolute error is not the 
most serviceable measure of the precision of a measurement. 
The relative error, which is defined as the ratio of the absolute 
error to the exact value, is often found more serviceable. Since 
the relative error is a ratio, it is an abstract number, and is 
therefore sometimes expressed in per cent. For example, if 
the diagonal of a square 10 in. on a side be measured and 
and found to be 14.1 in., the absolute error is less than 1/10 
of an inch. The relative error is less than (1/10) -^ 10 \/2 
= 1/141, approximately, le. less than 0.71 per cent. 

158. Rounded Numbers. Significant Figures. When the 
result of a measurement is expressed in the decimal notation, 
a generally adopted convention makes it possible to determine 
the degree of precision of the measurement from the number 
of significant figures contained in the number expressing the 
measure. This convention simply specifies that no more digits 
shall be written than are (probably) correct. Thus a measure- 

* The absolute error is therefore positive or negative according as the ob- 
served value is too small or too large. 

236 



IX, § 158] NUMERICAL COMPUTATION 237 

ment of a length, expressed as 14.1 in. means that the measure 
is exact to the nearest tenth of an inch. If on the other hand 
the measurefment of this length were exact to the nearest 
hundredth of an inch, the measure would have been expressed 
by the number 14.10.* 

We should note, then, that the two numbers 14.1 and 14.10 
do not mean precisely the same thing, when they express the 
result of a measurement. 

Again we may note that the absolute errors involved in the 
expression 4371.52 ft. and 42.81 ft. are each less than one 
hundredth of a foot ; whereas the relative error is in the first 
case less than 1/437152 and in the second only less than 
1/4281. 

Sometimes we are furnished with numbers expressing meas- 
ures which are given with greater accuracy than we can use, or 
care to use. Thus suppose we want to express a measured 
length of 3.5 in. in terms of centimeters. We find in a table 
of equivalent lengths that 1 in. = 2.54001 cm. It would be 
obviously absurd to use this expression as it stands. We 
accordingly round it off to 2.54 or even to 2.5 and find that 3,5 
in. = 8.9 cm. If, on the other hand, we wish to express 3.50000 
in. in centimeters, we should have to use the value 2.54001. 

A number is rounded off by dropping one or more digits at 
the right, and, if the last digit dropped is 5% 6, 7, 8, or 9, in- 
creasing the preceding digit by l.f Thus the successive 
approximations to tt obtained by rounding off 3.14159 .- are 
3.1416, 3.142, 3.14, 3.1, 3. 

* In other words x = 14.1 means that the exact value of x lies between 14.05 
and 14.15; and x = 14.10 means that the exact value of x lies between 14.095 
and 14.105. 

t In rounding off a 6, computers use the following rule : Always round off 
a 6 to an even digit. Thus 1.415 would be rounded to 1.42, whereas 1.445 
would be rounded to 1.44. The reason for this rule is that, if used con- 
sistently, the errors made will in the long run compensate each other. 



238 MATHEMATICAL ANALYSIS [IX, § 158 

The significant figures of a number may now be defined as 
the digits 1, 2, 3, -, 9 together with such zeros as occur 
between them or as have been properly retained in rounding 
them off. Thus 34.96 and 3,496,000 are both numbers of four 
significant figures. On the other hand 3,496,000.0 has eight 
significant figures, since the in the first decimal place accord- 
ing to the convention adopted means that the number is exact 
to the nearest tenth. This zero is then essentially a digit 
properly retained in rounding off, and should be counted as one 
of the significant figures. 

Confusion can arise in only one case. For example, if the 
number 3999.7 were rounded by dropping the 7, we should 
write it as 4000 which, according to the rule just given, we 
would consider as having only one significant figure, whereas in 
reality we know from the way in- which the number was ob- 
tained that all four of the figures are significant. When such 
a case arises in practice we may simply remember the fact, or 
we can indicate that the zeros are significant by underscoring 
them, or by some other device. Computers adopt devices of 
their own to avoid errors in such cases. 

159. Computation with Rounded Numbers. Addition. 

Since the (absolute) error of any approximate number can be 
at most one half the unit represented by the last digit at the 
right, the sum of n such numbers can be in error by at most 
n/2 times the unit represented by the last figure. These con- 
siderations lead to the following convention : in adding a 
column of approximate numbers first round off the given 
numbers so that not more than one column at the right is 
broken ; round off the sum so that the last figure to the right 
comes in the last unbroken column. This last figure is then 
uncertain. Nevertheless, it is usually retained temporarily. 
As a matter of fact, even the figure preceding this last one is 



IX, § 160] NUMERICAL COMPUTATION 239 

not certain, since the errors may accumulate in adding several 
numbers. 

Eor example, in adding 21.64 

3.8576 
5.259743 
10.31 

we first round off : 21.64 

3.858 

5.260 
10.31 
41.068 = 41.07 

The final sum is written 41.07, but even the last figure 7 is 
open to question. Show that the true result may be as low as 
41.06 or as high as 41 .08. 

To retain all the figures in the second and third of the num- 
bers originally given would be absurd and would give in the 
result a misleading pretense of accuracy which does not exist 
in fact. 

In subtraction round off similarly. 

160. Multiplication. Let a and b be approximate numbers 
and let their relative errors be a and ft respectively. The 
exact numbers are then (nearly) a -{- aa and b + bft. Their 
product is a6 + «6«+a&;8 + aJ«j8. 

The error committed in using ab as the product is then 
ab{a + ^ + «^) 

and the relative error is therefore nearly 

Now in practice a and /3 are small fractions, so that a ft is in- 
significant when compared with a + /8. (For example, if a and 



240 MATHEMATICAL ANALYSIS [IX, § 160 

/3 are both equal approximately to 0.001, ap is equal approxi- 
mately to 0.000001.) Hence, we conclude that the relative 
error of the product of two numbers is approximately equal to the 
sum of the relative errors of the factors. 

Hence, in finding the product of two approximate numbers, 
round off so that the two numbers have the same number of 
significant figures, and retain only this number of significant 
figures in the product* Even then the last figure retained may 
be unreliable. 

Example. Multiply 27.17 by^3, 14159. Round off the second factor to 
3.142, and multiply: 

27.17 X 3.142 

5434 

10868 

2717 
8151 
85.36814 = 85.37 

Even the figure 7 may be in error. Show that the true answer may be as 
low as 85.35. 

The labor involved in such a multiplication may be considerably re- 
duced by slightly modifying the method used, as follows : 

After having equalized the number of significant fig- 
ures annex a zero to the multiplicand. Multiply by the 
first figure on the left of the multiplier. Drop the last 
figure of the multiplicand and multiply by the second 
figure of the multiplier. Drop the next figure of the 
multiplicand and multiply by the third figure of the 
multiplier (but "carry" the amount from the figure 
dropped : thus .in the example having dropped the 7 and multiplying by 4, 
we say 4 x 7 = 28, carry 3, 4 x 1 = 4, +3 = 7, which is the first figure 
we write), and so on, arranging all the partial products so that the last 
figures from the left fail into the same vertical column ; then add in the 
usual way. 

* Since ^^ = 3.1428571, while tt = 3.1415927, the value y may be used for ir 
when the uncertainty of the other factors in a product in which it appears 
is greater than 1 part in 3000 (approximately) . 



27.170 X 


3.142 


81510 




2717 




1087 




64 




85368 = 


: 85.37 



IX, § 161] NUMERICAL COMPUTATION 241 

161. Division. In case either the dividend (iV) or the 
divisor (D) is an approximate number, the following shortened 
method may be used : 

1. Equalize the relative accuracy of N and D ; but if D is 
larger at the left, keep one extra figure on JV (as in the example 
below). 

2. Divide as in long division, but drop successive figures in 
Z>, instead of adding successive zeros to N. 

Example. Find 295.679 -r- 7.53. (As 7 is greater than 2, we retain 
four figures in the dividend.) 

7.53 1 295.7 [39.3 

225 9 [3 X 753] 

69 8 [divide by 75, gives 9] 

67 8 [9x3 = 27, carry 3 ; 9 x 75 = 675, + 3 = 678] 

2 [divide by 7, gives 3 (nearer than 2)] 

EXERCISES 

1. Add the following numbers, each representing the result of a meas- 
urement: 25.62, 341.718, 2.62394, 28.7125. 

2. Express 5.216 inches in centimeters. 

3. Express 53.291 cm. in inches. 

4. A rectangular table top is measured, and is found to be 
2'4".5 X 3'6".4. Find its area. Find the error caused in this area if the 
measurements are each O'M too short. Find the relative error in the 
area. 

5. Assuming that you can estimate the length and the breadth of a 
room which is about 15' by 18' to within 2', how nearly can you estimate 
its area ? 

6. Assuming that you can measure each of the dimensions of the room 
of Ex. 5 with a yardstick to within 1" error, how nearly can you find the 
area of the floor ? If the height of the room is about 10', how nearly can 
you find the volume of the room by measurement ? 

7. Assuming that you can measure the radius of a circle about 6" in 
diameter to within 0".l error, how nearly can you find its area ? How 
nearly could you find by measurement the volume of a cylinder about 5' 
high and about 5" in diameter ? 



242 MATHEMATICAL ANALYSIS [IX, § 162 

II. LOGARITHMIC SOLUTION OF TRIANGLES 

162. Logarithmic Computation. We have already had 
occasion to observe that many computations in engineering, 
astronomy, etc., are carried out by means of logarithms. Jn 
the last chapter a few examples of the use of logarithms in 
computation were given in connection with a four-place table. 
Such a table suffices for data and results accurate to four sig- 
nificant figures. When greater accuracy is desired we use a 
five-, six-, or seven-place table. 

The methods used in connection with such a table differ 
slightly from those used ordinarily with a four-place table. 
Accordingly we take up briefly at this point some problems in- 
volving computation with a five-place table of logarithms. 

No subject is better adapted to illustrate the use of logarith- 
mic computation than the solution of triangles, which we shall 
consider in some detail. Five-place tables and logarithmic 
solutions ordinarily are used at the same time, since both tend 
toward greater speed and accuracy. 

163. Five-place Tables of Logarithms and Trigonometric 
Functions. The use of a five-place table of logarithms differs 
from that of a four-place table in the general use of so-called 
" interpolation tables " or " tables of proportional parts," to facil- 
itate interpolation. Since the use of such tables of proportional 
parts is fully explained in every good set of tables, it is unnec- 
essary to give such an explanation here. It will be assumed 
that the student has made himself familiar with their use.* 

In the logarithmic solution of a triangle we nearly always 
need to find the logarithms of certain trigonometric functions. 

* For this chapter, such a five-place table should be purchased. See, for ex- 
ample, The Macmillan Tables, which contain all the tables mentioned here 
with an explanation of their use. 



IX, § 163] NUMERICAL COMPUTATION 243 

For example, if the angles A and B and the side a are given, 
we find the side b from the law of sines given in § 125, 

, a sin B ' 
b = — — -' 

sin A 

To use logarithms we should then have to find log a, log (sin B) 
and log (sin A). With only a table of natural functions and a 
table of logarithms at our disposal, we should have to find first 
sin Ay and then log sin A. For example, if A = 36° 20', 
we would find sin 36° 20' = 0.59248, and from this would find 
log sin 36° 20' = log 0.59248 = 9.77268 - 10. This double use 
of tables has been made unnecessary by the direct tabulation 
of the logarithms of the trigonometric functions in terms of 
the angles. Such tables are called tables of logarithmic sines, 
logarithmic cosines, etc. Their use is explained in any good 
set of tables. 

The following exercises are for the purpose of familiarizing 
the student with the use of such tables. 

EXERCISES 

1. Find the following logarithms : * 

(a) log cos 27° 40'.5. (d) log ctn 86° 53'. 6. 

(6) log tan 85° 20'. 2. (e) log cos 87° 6'. 2. 

(c) log sin 45° 40'.7. (/ ) log cos 36° 53'.3. 



(d) log sin A = 9,78332 - 10. 

(e) log ctn ^A = 0.70352. 
(/) log tan J ^ = 9.94365 - 10. 



87325 

4. Given a triangle ABC, in which ZA = 32°, ZB = 27°, a - 5.2, find 
b by use of logarithms. 

* Five-place logarltbms are properly used when angles are measured to the 
nearest tenth of a minute. For accuracy to the nearest second, six places 
should be used. 



2. Find A 


, when 






(a) log sin 


A = 9.81632 - 


-10. 




(6) log cos 


A = 9.97970 - 


-10. 




(c) log tan ^ = 0.45704. 






3. Find^, 


iftan. = 4^«-\2^ 

Q7Q. 


89.710 



244 MATHEMATICAL ANALYSIS [IX, § 164 

164. The Logarithmic Solution of Triangles. The effective 
use of logarithms in numerical computation depends largely on a 
proper arrangement of the work. In order to secure this, the 
arrangement should be carefully planned beforehand by con- 
structing a blank form, which is afterwards filled in. Moreover 
a practical computation is not complete until its accuracy has 
been checked. The blank form should provide also for a good 
check. Most computers find it advantageous to arrange the 
work in two columns, the one at the left containing the given 
numbers and the computed results, the one on the right contain- 
ing the logarithms of the numbers each in the same horizontal 
line with its number. The work should be so arranged that 
every number or logarithm that appears is properly labeled ; 
for it often happens that the same number or logarithm is used 
several times in the same computation and it should be possible 
to locate it at a glance when it is wanted. 

The solution of triangles may be conveniently classified 
under four cases : 

Case I. Given two angles and one side. 

Case II. Given two sides and the angle opposite one of the 
sides. 

Case III. Given two sides and the included angle. 

Case IV. Given the three sides. 

In each case it is desirable (1) to draw a figure representing 
the triangle to be solved with sufficient accuracy to serve as a 
rough check on the results ; (2) to write out all the formulas 
needed for the solution and the check ; (3) to prepare a blank 
form for the logarithmic solution on the basis of these 
formulas ; (4) to fill in the blank form and thus to complete 
the solution. 

We give a sample of a blank form under Case I ; the student 
should prepare his own forms for the other cases. 



IX, § 165] NUMERICAL COMPUTATION 



245 



165. Case I. Given two Angles and one Side. 

Example. Given : a=430.17, ^=47° 13'.2, B=d2° 29'.6. (Fig. 134) 
To find: C, 6, c. 
Formulas : 



6 = 



180°- 

a 

sin J. 

a 



-(A+B), 
sin B, 

sin C. 



Check : 




Fia. 134 



sin^ 
c-b ^ ts,ni(C-B) 
c + b tSinl{C-\-B)' 

The following is a convenient blank form for the logarithmic solu- 
tion. The sign ( + ) indicates that the numbers should be added; the 
sign ( — ) indicates that the number should be subtracted from the one just 

above it. ^ , , 

Logarithms 



A- 


Numbers 


( + )^ = 










A + B = 
C = 








179° 60' 


d — 






sin^ = 


sin 


• . 


a/sin A 






sin B = 
b = 


sin 


• • 






•) (+) 

■) 



a /sin A 

sin C = sin 
c =.. . 



c-b = 
c+ b = 



(->) ( + ) 



Check 



C-B= 

C + B = . .... 

tan^C- J5) =tan . . . (- 
tan^(C-|-5)=tan . . . (- 



■(1) 

(Logs (1) and (2) 
. should be equal 



.) ( — ) for check.) 

(2) 



246 



MATHEMATICAL ANALYSIS 



[IX, § 165 



FilUng in this blank form, we obtain the solution as follows. 

Numbers Logarithms 

A = 47° 13'.2 

B= 52° 29\6 

A-\-B= 99°42'.8 

179° 60'.0 



C = 80° 17'.2 

a = 430.17 (->) 2.63364 

sin ^ = sin 47° 13' -2 (— >) (-) 9.86567-10 

a/sin ^ 2.76797 

sin B = sin 52° 29'. 6 (->) ( + ) 9.89943 - 10 

b = 464.94 Ans. (<-) 2.66740 

a/sin yl 2.76797 

sin C = sin80° 17'.2 (->) ( + ) 9.99373 

c = 577.70 Ans. (^^) 2.76170 



c-6= 112.76 
c + 6 = 1042.64 



(->) 2.05215 

(->►) (-) 3.01818 

9.03402 - 10 



C-B= 27°47'.6 

C + 5 = 132° 46'.8 
tan ^ ( a - jB) = tan 13° 53'.8 (->) 9.39342 - 1 

tan 1(0+ -B)= tan 66° 23'. 4 (— ^) (_) 0.35942 

9.03400 - 10 J 



Check 



EXERCISES 

Solve and check the following triangles ABC : 

1. rt = 372.5, .4 = 25°30', J5 = 47°60'. 

2. c = 327.85, vl = 110° 52'.9, B = 40° 31'.7. Ans. C = 28° 35'.4 

a = 640.11, b =446.20. 

3. a = 53.276, A = 108° 50'.0, C = 57° 13'.2. 

4. 6 = 22.766, ^=141°59'.l, a=25°12'.4. 

5. 6 = 1000.0, B = 30° 30'. 5, C = 50° 50'.8. 

6. a = 257.7, A = 47° 25', B = 32° 26'. 



IX, § 166] NUMERICAL COMPUTATION 



247 



166. Case II. Given two Sides and an Angle opposite 
one of them. 

If A, a, b are given, B may be determined from the relation 

h sin A 



(1) 



sin B 



If log sin B = 0, the triangle is a right triangle. Why ? 

If log sin B >0, the triangle is impossible. Why ? 

If log sin B <0, there are two possible values Bi, B^ of B, 
which are supplementary. 

Hence there may be two solutions of the triangle. (See Ex. 1, 
page 249.) 

No confusion need arise from the various possibilities if the 
corresponding figure is constructed and kept in mind. 

It is desirable to go through the computation for log sin B 
before making out the rest of the blank form, unless the data 
obviously show what the conditions of the problem actually 
are. 

Example 1. Given : A = 46° 22'.2, a = 1.4063, 5=2.1048. (Fig. 135) 
To find: S, O, c. 



Formula : sin B 



b sin A 
a 




Fig. 135 



Numbers 
& = 2.1048 (- 

sin ^ = sin 46° 22'. 2 (- 
bsinA 

a = 1.4063 (- 

sin B {< 



Logarithms 
0.32321 

►) ( + ) 9.85962 ~ 10 

0.18283 
(-) 0.14808 
.) 0.03475 



Hence the triangle is impossible. Why ? 



248 



MATHEMATICAL ANALYSIS 



[IX. § 166 



Example 2. 



Given : a = 73.221, b = 101.53, A ■. 
To find: J5, C, c. 



40° 22'.3. (Fig. 136) 



Formula: sin 5 = 



6sin^ 



Numbers 
b = 101.63 ( 

sin^=sin40°22'.3 ( 
b sin A 

a = 73.221 
sin JB 

G 



Logarithms 
2.00660 

-) ( + ) 9.81140 -10 
11.81800 - 10 
1.86464 



(-^) (-) 



9.95336 - 10 




The triangle is therefore possible and 
has two solutions (as the figure shows). 
We then proceed with the solution as 
follows : 

We find one value Bi of B from 
the value of log sin B. The other 





L JJ lO llUCJU j 


Other formulas : 






(7=180°-(^ + B). . 




asin C 
sin^ 






Check: ^-^: 
c + & 


tanUC-B) 
tanKO + 5) 


Numbers 




Logarithms 


sin B 




9.95336 - 10 


Bi= 63°65'.2 






179*^ 60'.0 






^2 = 116° 4'.8 






-4 + 5i = 104° 17'.6 






179° 60'.0 






Ci= 76'^42'.6 






a 


(-» 


1.86464 


a'mA 


(-» (- 


-) 9.81140- 


a/sin A 




. 2.06324 


sinCir=8in76°42.'6 


'(->) ( + ) 9.98634-] 



Cl 



10 

10 

109.64 (-^) 2.03968 



IX, § 167] NUMERICAL COMPUTATION 



249 



Ci - 6 = 8.01 (. 

Ci + 6 = 211.07 (. 

Ci-^i= 11°47'.8 
Ci + ^i = 139°37'.7 
tan K Ci - Bi) = tan 6° 63'.6 
tan K Ci + Bi) = tan 69=^ 48'.8 



0.90363 

.) (-) 2.82443 

8.57920-10 



9.01377 - 10 
0.43455 
8.57922 - 10 



Check* 



One solution of the triangle gives, therefore, B=63° 55'.2, C = 76° 42'.5, 
c = 109.54. 

To obtain the second solution, we begin with B2 = 116° 4'. 8. We find 
C2 from C2 = 180° -(A + B^) ; i.e. C% = 2-3° 32'. 9. The rest of the com- 
putation is similar to that above and is left as an exercise. 



EXERCISES 

1. Show that, given A, o, 6, if A is obtuse, or if A is acute and a > 6, 
there cannot be more than one solution. 

Solve the following triangles and check the solutions : 

2. a = 32.479, 6 = 40.176, ^ = 37° 25M. 

3. 6 = 4168.2, c = 3179.8, B = 51° 21'.4. 

4. a = 2.4621, b = 4.1347, B = 101° 37'.3. 

5. a = 421.6, c = 532.7, ^ = 49° 21'.8. 

6. a = 461.5, c= 121.2, C = 22° 31'.6. 

7. Find the areas of the triangles in Exs. 2-6. 

167. Case III. Given two Sides and the Included Angle. 

Example. Given: a=214.17, 6=356.21, 
C=62°21'.4. (Fig. 137) 

To find : A, B, c. 
Formulas : 
6 



tan I {B- A) 



tan KB + A); 



b + a 
B+A= 180° - C = 117° 38'.6 ; 
_ gsin C _ 6 sin C 
sin A sin B 

* A small discrepancy in the last figure need npt cause concern. Why? 




250 



MATHEMATICAL ANALYSIS 



[IX, § 167 



Numbers 
6 - a = 142.04 ( 

6 + a = 570.38 ( 

(b - a)/{h + a) 

tan ^{B + A) = tan 58° 49' .3 ( 
tan \{B- A) = tan 22° 22^2 ( 
.•.A= 36°27M 
5= 81°11'.5 





-) 
-) 

Ans. 

Ans. 

a = 214.17 (->) 

sin ^ = sin 36° 27'. 1 (->) 
a/sin ^ 

sin C = sin 62° 21 '.4 (->) 
c = 319.32 Ans. (^ 
Check by finding log (6/sin B). 



Logarithms 

2.16241 
(-) 2.75616 

9.39625 - 10 
( + ) 0.21817 

9.61442-10 



2.33076 
(-) 9.77389 -10 

2.55687 
( + ) 9.94736 -10 

2.50423 



EXERCISES 

Solve and check each of the following triangles. 

1. a = 74.801, h = 37.502, C = 63° 35'.5. 

2. a = 423.84, 6 = 350.11, C = 43°14'.7. 

3. 6 = 275, c = 315, ^ = 30° 30'. 

4. « = 150.17, c = 251.09, ^ = 40°40'.2. 

6. a = 0.25089, b = 0.30007, C = 42° 30' 20". 
6. Find the areas of the triangles in Exs. 1-5. 



168. Case IV. Given the three 
Sides. 

Example. Given : a = 261.62, 
b = 322.42, 
c = 291.48. 
To find : A, B, G. 
Formulas : 

s = l{a -\- b -\- c). 
y-J (« — q)(g-ft)(g-c) 



tan ^A= — ^ , tan I 5 = ** 



a s — b 

Check: A -\- B -\- C = 180°. 




tan I C = 



b=3gg.4g A 

Fig. 138 



8 — C 



(§ 143) 



IX, § 168] NUMERICAL COMPUTATION 251 



Numbers 
a = 261.62 
6 = 322.42 
c = 291.48 


( + ) 
(-) 


Logarithms 


2s = 875.52 
s = 437.76 
s-a= 176.14 (->) 
s-b= 115.34 (-^) 
s-c = 146.28 (-^) 


2.24586 
2.06i98 
2.16518 


2s = 875.52 (Check.) 
8 = 437.76 (->) 


6.47302 
2.64124 


r 
s— a 




3.83178 
1.91589 
2.24586 


tan ^ ^ = tan 25° 4'.1 (<-) 




9.67003 - 10 


r 

s-b 




1.91589 
2.06198 


tan 1 5 = tan 35° 32'.4 (<— ) 
r = 
s— c = 




9.85391-10 

1.91589 

2.16518 


tan ^ C = tan 29° 23'. 4+ (<-) 
A = 50° 8'. 2 ^ws. 
B= 71° 4'.8 ^ns. 
0= 68° 46'. 9 Ans. 




9.75071-10 



179° 59'. 9 Check. 

EXERCISES 

Solve and check each of the following triangles : 

1. a = 2.4169, b = 3.2417, c = 4.6293. 

2. a = 21.637, & = 10.429, c = 14.221. 

3. a = 528.62, b = 499.82, c = 321.77. 

4. a = 2179.1, & = 3467.0, c = 5061.8. 

5. a = 0.1214, 6=0.0961, c = 0.1573. 

6. Find the areas of the triangles in Exs. 1-5. 

7. Find the areas of the inscribed circles of the triangles in Ex. 1-6. 



252 



MATHEMATICAL ANALYSIS [EX, § 169 



III. THE LOGARITHMIC SCALE — THE SLIDE RULE 

169. The Logarithmic Scale. Let us lay off, on a straight 
line, segments issuing from the same origin and proportional 
to the logarithms of the numbers 1, 2, 3, 4, —. The base of 
the system of logarithms is immaterial. Let us label the end- 
points of these segments by the corresponding numbers. This 
gives a non-uniform scale, which is called a logarithmic scale. 
Such a scale is pictured in Fig. 139. 



[ 



T 



s 4 

Fig. 139 



rm 



A scale of this kind is easily constructed from the graph of 
the logarithmic function (Eig. 133). 

170. The Slide Rule. The slide rule is an instrument often 
used by engineers and others who do much computing.* It 
consists of a rule (usually made of wood faced with celluloid) 




Fig. 140 

along the center of which a slip of the same material slides 
in a groove. This slip is called the slide. The face of the 
slide is level with the face of the rule. 

* Engineers usually purchase rather expensive slide rules made of wood 
and celluloid. These are on sale in all stores which carry draftsmen's supplies. 
A very simple slide rule sufficiently accurate for class purposes is printed on 
hard pasteboard and is obtainable at reasonably small cost through any one 
of several manufacturers of instruments. Figure 140 is reproduced on a larger 
scale on the first fly-leaf at the back of the book. By cutting out this leaf 
and carefully cutting up the figure, a slide rule can be made by the student. 
This will not be very accurate, but it will suffice to illustrate the principles. 



XI, § 170] NUMERICAL COMPUTATION 



253 



Along the upper edge of the groove are engraved two loga- 
rithmic scales, usually labeled A and B, the scale A being on 
the rule, the scale B on the slide. (See Fig. 141.) 

The scales A and B are identical. The slide is simply a 
mechanical device for adding graphically the. segments on 



r 

1 2 


C 




I 5 6 7 8 9 1 

^ 1 I 1 1 I 1 , , , , 1 




2 


^ ihiililijlilililiilililili 


[llllili 




m 


m 


m 


w'm 


T-n: 


■] 


trhl 




1 


\ f v> 


1 1 


1 


m 


lU 


ill 


1 p 


M 1 1 


:::nn]iiioi 




1 


\ 


C 

N 


t-n , ,,,,, ...^T-g; ■ 4 ■ , 


) 6 


7 8 9 


I 


2 


3 


y 


iTi"TiiiT-r 


'IHW 


]]Xii"":i 


"■""I""I] 


^iiiiiiiiiiiiiiiiiiiiiiiiiii 


U.iuM.-ljLi 


' 1 


..Mr.i:ii 


:ii:i4 


1 2 


J 


4 



Fig. 141 

these scales. Since the segments represent the logarithms of 
the numbers found on the scale, the operation of adding the 
segments is equivalent to multiplying the corresponding num- 
bers. Thus, to find the product 2.5 x 3.2 move the slide to the 
right until the point marked 1 at the extreme left of the 
slide (scale B) is in contact with the point 2.5 on scale A 
(Fig. 141 shows the positions of scales A and B after this 
operation). The point 3.2 on scale B is then opposite the point 
8.0 on scale A. The latter number is the required product : 
2.5 X 3.2 = 8.0. A little reflection should make quite clear 
how the operation just performed is equivalent to adding the 
logarithms of 2.5 and 3.2 and then reading from the scale the 
number corresponding to the sum. We may note further that 
with slide set as in the example just worked it is set for 
multiplying any number by 2.5 ; i.e. every number of the scale 
A is the product of 2.5 by the number below it on scale B. 
The slide is therefore also set for division by 2.5. Every 



254 MATHEMATICAL ANALYSIS [IX, § 170 

number of scale B is the result of dividing the number above 
it by 2.5. Thus we read from the scale (set as before) that 
7.2 -^ 2.5 = 2.9 approximately. 

Having now shown very briefly how the slide rule may be 
used for multiplication and division, let us examine it a little 
more closely. Scales A and B are labeled with the numbers 

1,2, 3, 4, 5, 6, 7, 8, 9, 1,2, ...,9,1. 
It is natural to ask why the number following the 9 in the 
middle of these scales is not labeled 10 ? The answer is that 
the numbers on the slide rule are given without any reference 
to the position of the decimal point, just as the numbers in a 
table of logarithms are given without reference to the decimal 
point. The number 1 at the extreme left of the scale may 
represent either 1, or 10, or 100, or 1000, etc., or .1, or .01, or 
.001, etc. If the 1 at the extreme left of the scale represents 
1, then the other numbers on the first half of the scale repre- 
sent 2, 3, ..., 9, the 1 in the middle represents 10, the 2 represents 
20, and the successive numbers represent 30, 40, -., 100 (the 
last being represented by the 1 at the extreme right of the 
scale). If on the other hand the 1 at the left represents 100, 
the successive numbers represent 200, 300, ..., 900, 1000, 2000, 
..., 10,000. If the 1 at the left represents .1, the successive 
numbers represent .2, .3, .••, .9, 1.0, 2.0, ••., 10.0 ; and so on. 

The reading of the subdivisions on the scales (A and B) 
should now offer little difficulty. Whenever an interval be- 
tween two successive numbers is divided by certain lines of the 
same length into 10 parts, each of these parts represents one 
tenth of the number represented by the interval in question. 
Thus, if we fix our attention on the division between 2 and 3, 
we note that a certain set of lines divides this interval into 10 
parts ; if the 2 represent 2, these divisions represent respec- 
tively 2.1,2.2, ..., 2.9. On the other hand, if the 2 is thought 



IX, § 170] NUMERICAL COMPUTATION 255 

of as representing 20, these divisions represent 21, 22, —, 29 : 
and so on. These divisions into ten are at some parts of the 
scale subdivided further into five or two parts. These parts 
then represent fifths or halves of the interval that represented 
a tenth. Thus we may readily locate on the scale the point 
representing 1.42 or the point representing 3.65. 

Turning our attention to scales C and D along the lower 
edge of the groove on the slide and the rule respectively, 
we note first that these two scales are also identical. Compar- 
ing them with scales A and B, we see that the unit chosen for 
C and D is just twice the unit of A and B. Hence the scales 
C and D can be used for multiplying and dividing just as 
scales A and B are used ; however on C and D our range is 
smaller. The range of numbers on A and B is from 1 to 100 ; 
on C and D only from 1 to 10. To make up for this limitation, 
scales C and D give greater accuracy. 

However, the principal reason for the existence of the second 
pair of scales is the fact that the two pairs of scales thus ob- 
tained furnish a table of squares and square roots. In view of 
the relation between the units with respect to which the two 
pairs of scales are constructed, every number of scale A is the 
square of the number vertically below it on scale D. Why ? 
In order that corresponding numbers on scales A and D may be 
accurately read off, every slide rule is provided with a runner, 
the vertical line on which connects corresponding numbers of 
the upper and lower scales. The runner also enables us to 
perform calculations consisting of several operations without 
reading off the intermediate results, thus saving time and 
securing greater accuracy in the final result. The actual use 
of the slide rule will be explained in the next article. 

The successful use of the slide rule depends largely on the 
ability to read the scales readily and accurately, accuracy 



256 MATHEMATICAL ANALYSIS [IX, §170 

often necessitating the estimating of numbers falling between 
the lines of division. The ability mentioned can be secured 
only by practice. A proficient operator, with a ten-inch slide 
rule, can always secure results accurate to three significant 
figures. This degree of accuracy is sufScient for many of the 
computations of applied science, manufacturing, etc., in which 
the slide rule is proving more and more useful. 

171. The Use of the Slide Rule. All calculations in mul- 
tiplication, division, proportion, etc., are worked on scales Cand 
D unless the answer is so large that it does not lie on the scale. 
In that case scales A and B are used. Let us begin with pro- 
portion. On this topic, and on the corresponding property of 
the slide rule, all computations involving multiplication or 
division, or both, maybe made to depend in a very simple way. 

The property of the slide rule referred to is as follows : No 
matter where the slide be placed, all the numbers on the slide 
bear the same ratio to the corresjjonding numbers on the rule (due 
regard being had to the position of the decimal point). For 
example, if the slide be set so that 2 of O coincides with 4 of 
D, it will be observed that the same ratio 2 : 4 exists between 
every pair of corresponding numbers : 1 : 2, 3 : 6, 42 : 84, 
125 : 250, etc. Explain why this is true. This leads at once to 
the rule for finding the fourth term of a proportion, when the 
first three are given. We give this rule in diagrammatic form, 
as follows : * 

To find the fourth term of a proportion : 



c 

D 


Set first term 
over second term. 


Under third term 
find fourth term. 



♦ Iji this article we have followed to a considerable extent the treatment 
given in the Manual for the use of the Mannheim Slide Rule, published by 
the Keuffel and Esser Co., New York. 



IX, § 171] NUMERICAL COMPUTATION 

This gives the solution of the equation 

b X 
To find the product abj solve the proportion 

a X 
To find the quotient -, solve the proportion 
a__x 

The following examples will make clear the procedure. 
Example 1. Solve the proportion : 13/24 = 32/a;. 



257 




D 


Set 13 
over 24 


Under 32 

find 69.1 Ans. 



Example 2. Solve the proportion : 13/24 = 75/x. 

Since the first two terms of the proportion are the same as in the pre- 
ceding example, we set the slide as before. We now find, however, that 
75 on C is beyond the extremity of D. We accordingly set the runner on 
the left-hand 1 of 0, and then set the right-hand 1 of C on the runner. 
We find under 75 the number 138.5, the required value of x* (Justify 
the above use of the runner. ) 

The same example can be done on scales A and B with one setting, 
without using the runner. 

Example 3. Find the product: 23.2 x 5.3. 



c 

D 


Setl 
over 23.2 


Under 5.3 

find 123.0 Ans. 



Here we set the right-hand 1 on 23.2. Use whichever 1 serves. The 
decimal point, in this as in the other examples, is simply located by in- 
spection and a brief mental estimate of the answer. Here we see readily 
that the answer is something over 100; hence we locate the decimal 
point at the place to give us 123.0. 

* The .5 in this answer must be estimated. Usually, if more than three 
significant figures are obtained from the rule, the last is uncertain. 



258 MATHEMATICAL ANALYSIS [IX, § 171 

Example 4. Find the value o/364 -4- 115. 



c 

D 


Set 364 
over 115 


Find 3.17, Ans. 
over 1 



Example 5. Find the circumference of a circle whose diameter is 42 
ft. We multiply the diameter by tt = 3.14.* Hence, 



c 

D 


Set 1 
over 3.14 


Under 42 

find 132.0 Ans. 



By ordinary multiplication we get 131.88 ; an example of the inaccur- 
acy of the fourth significant figure. 

Example 6. Find the continued product : 1.6 x 4.2 x 5.3 x 2,8. 
The abbreviation R. denotes the runner on the slide-rule. 



Set 1 
over 1.6 



R, to 4.2 



1 to R. 



R. to 5.3 



1 toR. 



Under 2.8 
find 99.7 Ans. 



We add a fevnr more rules for computing various types of expressions 
involving scales A and B as vv^ell as C and D. 

(1) To find a^ xb: 



(2) To, 



A 




Find a2&. 


Ans. 


B 




over b. 




C 


Setl 






D 


over a 






ada'^-h 


b: 






A 




Find a2 h- b, 


Ans 


B 


Set 6 


over 1. 




C 








D 


over a 







* The number t is usually raa>'ked on the scale. 



IX, § 171] NUMERICAL COMPUTATION 



259 



(3) To find geometric mean between two numbers a and b; i.e. lincl x, 
so that a/x = x/b. Let a < 6. 



A 






B 


Seta 


Below 6 


C 






D 


over a 


findiX=G.M: 



(4) To reduce fractions to decimals 



Set numerator 
over denominator 



Find equivalent decimal 
above 1 



These rules are not to be memorized. They will be used almost in- 
stinctively by one who has made the reason for each rule thoroughly clear 
to himself and who is in practice. 



EXERCISES 

1. With a slide rule compute the value of : 

(a) 2.13 X 4.42. {h) 2,856,000 x 256,700,000. 

(&) 1.98x5.24., ___ 5,43^31.5 



(c) 2.77 x 3.14 X 4.25. 
id) 8.27/2.63. 
(e) 5.48/3.26. 
(/) 10/3.14. 
{g) 0.000116 X 0.0392. 



(0 

U) 



21.4 

7.64 X 4.14 
21.2 

67.4 X 25.5 X 19.7 



4.64 X 18.4 

2. With a slide rule compute the value of : 

(a) (2.85)2. (c) (1.86)3. 

(6) 3.72 X (2.23)2. (^) (6.24)2/26.3. 

3. Find the circumference and the area of a circle whose radius is 
4.16 in. 

4. What is the length in feet of 27.3 meters, given that 26 meters = 
82 feet ? Solve with one setting of the slide. 



260 MATHEMATICAL ANALYSIS [IX. § 172 

IV. LOGARITHMIC PAPER 

172. Logarithmic Paper. Euled paper is printed, on which 
the rulings in both directions are spaced according to the 
logarithmic scale (§ 169), i.e. precisely as on a slide rule.* 
Such paper is called logarithmic paper. Samples of this ruling 
are shown in Figs. 142-143. 

173. Plotting Powers on Logarithmic Paper. The graphs 
of equations of the type 

(1) y = kx"" 

can be plotted very readily on logarithmic paper. For, if we 
take the logarithms of both sides, we find 

(2) log 2/ = log A: -h n log x. 

Let us set Y=\ogy, K=\ogk, X=loga;; 
then (2) becomes 

(3) Y=K-\-nX. 

Now the equation (3) represents a straight line if X and Y be 
taken as the variables. This is precisely what happens if we 
plot the values of x and y from equation (1) on logarithmic 
paper ; for, when we plot a value for x on logarithmic paper, the 
distance from the left border is nothing else than logic, i.e. X; 
and similarly for Y. 

Moreover, the slope of the straight line represented by (3) is 
n, the exponent of x in (1) ; and the intercept on the Y axis is 
K= log k. Hence if values of x and y from (1) are plotted on 
logarithmic paper, the value of n in (1) appears as the slope of 
the straight line graph, and the value of k can be read off 
directly on the vertical axis. 

* On this account, it is possible to make a crude slide rule by using the 
edges of two sheets of logarithmic paper, sliding them along each other after 
the manner of a slide rule. 



IX, § 173] NUMERICAL COMPUTATION 



261 



Example 1. Draw the graph of the equation y = r^ on logarithmic 
paper. 

Take x — \, then y =\. Take x = 10, then y = 100. Plot these two 
points A (1, 1) and B (10, 100) (Fig. 142). Connect A and 5 by a 
straight line. This is the required graph. 

The graph may be drawn also by noticing that its slope is the exponent 



ifm 


11,1111 1 1 1 1 1 1 11 


-y!:::::::::: -: 




o 


II II 1 


t 








2 ___ 




Q 








5 --- 


::::::: :: t 






/ 


::::::: :: i 








f 






3 -- 


7 








t 






.... 


7 

::::::: ^ 






1.5 

2 L 


::::::: ^E=- — 

::::::: izzz zi 


7 




9 


17 








z 






^ 


r^ 








::::::7 :: 








/ 






/ 










J 








f 














__ 


t 







'i 


_ 




La? 



Al 



1.5 2 



4 5 



7 S91 



1.5 



4 5 6 7 891 



Fig. 142 
of X in the given equation, i.e. 2. Hence we may draw from A a line 
whose slope is 2. Show that this gives the same line, AB. 

We may use this graph to find squares or square roots. Thus, if a; = 4, 
we can note the point on the graph directly over 4, and read the_corre- 
sponding value of y, which is 16. Reversal of the process gives \/16 = 4. 
Likewise, if x = 4.5, we find y - 20.2+ ; and vl5 = 3.8, approximately. 



262 



MATHEMATICAL ANALYSIS 



[IX, § 173 



Conversely, given a straight line on logarithmic paper, we 
know that its equation must be of the form (1). We can j&nd 
n by actually measuring the slope, and we can read off k on the 
vertical line through the point marked (1, 1), since if we place 



lOjYTTl 




— 










— 






7 


9 — 




















t 


n 




















7 " 






















/ 




















« 


^ 




















/ 




J. 


















/ 




■y 


" e= elongation in c 
i^r^pull in kg. 










y 






m. 








/ 












i 






2 


' e = .3 


r 










J 


f 







1.5 




^ 










^ 












' — 








--—< 


«J — 




















/ 























/ 






















_ TZ 






















r 










■y 






















c 












/ 






















/ 






















/ 




















I 






















~~T 






















k^-S,/__ 






















1 












.2 




= 










= 








.15 




E 






i^----------A 




E 


1 

■ 




1 










17 






















X 














.iLLLLL 




_J 










1 









,15 .2 



.3 



.4 .5 .6 .7.8.91 1.5 
Fig. 143 



3 4 5 6 7 8 910 



a; = 1 in equation (2), we have log?/ = logfc, whence 2/ = A;. 
Any other value of x may be used instead of £c = 1, but a; = 1 is 
most convenient because log 1 = 0. 

Example 2. A strong rubber band stretched under a pull of p kg. 
shows an elongation of e cm. The following values were found in an 
experiment : 



IX, § 173] NUMERICAL COMPUTATION 



263 



p 


0.5 


i:o 


1.5 


2.0 


2.5 


3.0 


3.5 


4.0 


4.5 


5.0 


6.0 


7.0 


e 


0.1 


0.3 


0.6 


0.9 


1.3 


1.7 


2.2 


2.7 


3.3 


3.9 


5.3 


6.9 



If these values are plotted on logarithmic paper, . it is evident 
that they lie reasonably near a straight line, such as that drawn in 
Fig. 143. 

By measurement in the figure, the slope of this line is found to be 1.6, 
approximately. Hence if we set 

P= logp, ^ = 'loge, 
we have ^=^+1.6P, 

where iTls a constant not yet determined ; whence 

loge = K-{- \.Q\ogp 
or . e = kp^-^^ 

where K = log k. If p = 1, e = A; ; from the figure, if j> = 1, e = 0.3 ; 

hence k = 0.3, and 

e = 0.3pi-6. 



EXERCISES 

1. Plot on logarithmic paper the graph of each of the following equa- 
tions : 

(a) y — 7?. (c) y = afi. (e) y = S x^. 

(b) y = xi (d) y = x^-s. (/) y = 4.5 x^-^. 

2. Draw the graph ot y = x-^. Note that the negative exponent — 2 
gives simply what we ordinarily call a negative slope of — 2 for the 
straight line graph. 

3. When air expands or is compressed (as in an air compressor) , with- 
out appreciable loss or gain of heat, the pressure p and the volume v are 
connected by the formula 

p = kv~^-^, approximately. 

Pressure is often measured in atmospheres, and volume in cubic feet. 
If we start with one cubic foot of air at one atmosphere of pressure, it is 
obvious that k = 1. Draw the graph for this case, and from it find p 
when V = 0.5 cu. ft. Find v when p = 5 atmospheres. Find v when 
p = 0.5 atmospheres. 



264 



MATHEMATICAL ANALYSIS [IX, § 173 



4. The intercollegiate track records for foot races (1916) are as 
follows, where d means the distance run, and t means the record time : 



cl 


100 yd. 


220 yd. 


440 yd. 


880 yd. 


Imi. 


2 mi. 


t 


0:09| 


0:21^ 


0:48 


l:54f 


4 : lof 


9:23f 



Plot the logarithms of these values on squared paper (or plot the given 
values themselves on logarithmic paper). Find a relation of the form 
t = k(P^. What should be the record time for a race of 1320 yd.? 

(See Kexnelly, Popular Science Monthly, Nov. 1908.) 

5. In each of the following tables, the quantities are the results of 
actual experiments ; the two variables are supposed theoretically to be 
connected by an equation of the form y = fccC*. Draw a logarithmic graph 
and determine k and ?i, approximately : 



(a) (Steam pressure ; v = volume, p = pressure. ) 



V 


2 


4 


6 


8 


10 


p 


68.7 


31.3 


19.8 


14.3 


11.3 



(Saxelby.) 



(6) (Gas engine mixture ; notation as above.) 



V 


3.54 


4.13 


4.73 


5.35 


5.94 


6.55 


7.14 


7.73 


8.05 


p 


141.3 


115 


95 


81.4 


71.2 


63.6 


54.6 


50.7 


45 



(Gibson.) 
(c) (Head of water h, and time t of discharge of a given amount.) 



h 


0.043 


0.057 


0.077 


0.095 


0.100 


t 


1260 


540 


275 


170 


138 



(Gibson.) 



CHAPTER X 

THE IMPLICIT QUADRATIC FUNCTIONS 

Two-valued Functions 

I. THE FORMS Ax^-^ Ey+ C = AND By'^ -{- Dx -{- C = 

174. The General Implicit Quadratic Function. We shall 
now return to the discussion of algebraic functions. We first 
discussed the explicit linear function y=mx-\-b, and the function 
y defined by the implicit relation Ax -f- By +0 = (Chapter 
III). Then we discussed the explicit quadratic function of 
the form y = ax"^ -\-hx -\- c (Chapter IV). We now propose to 
take up the discussion of the functions y defined by implicit 
quadratic relations, such as 4 !/2 — 5 a? = 0, ic^ — 4 ?/--f 2 a;— 41/— 1 
= 0, etc. The most general form of such an equation is 

(1) Ax"^ + Fxy + By^ + Dx -}- Ey + C = 0. 

The graphs of equations of this form are important curves, 
with interesting geometric properties, which we shall discuss 
in a later chapter. Our present purpose is to determine the 
general nature of these graphs (their shape, etc.) and to develop 
methods whereby the graph of a given equation of the type 
considered may be readily drawn. 

We may note at the outset that the function defined by an 
implicit quadratic relation between x and y will usually be 
two-valued, i.e. to each value of x will correspond, in general, 
two distinct values of y. This is due to the fact that if any 
particular value be assigned to x in equation (1) above, the 

265 



266 MATHEMATICAL ANALYSIS [X, § 174 

corresponding values of y are determined by a quadratic equa- 
tion, unless ^ = 0. 

We shall approach the discussion of equations of type (1) by 
considering in order certain simpler forms of this general 
type. First, we shall discuss equations of the two types 

Ax'+Ey +C = and By'^ ^ Dx -\- C =^ 0. 

175. The Equations x'^ — y — and y'^—x—0. We can dis- 
pose of the equations x"^ — y = and 3/2 — a; = very quickly. 
The first equation is equivalent to the equation y = x^, already 
discussed in § 72. The second equation is equivalent to the 
equation 

(2) y^=x, 

or y = ± Va;. 

We can either plot the points {x, y) whose coordinates satisfy 
this relation and thus obtain the graph desired * ; or, we can 
note that the equation y"^ = x is obtained from the equation 
352 = ?/ by simply interchanging x and y. Hence, the graph of 
y"^ = X is obtained from the graph of y = x^ by turning the 
plane of the graph oi y — x"^ over about the line through the 
origin bisecting the first and third quadrants. Eor, this opera- 
tion will interchange the x- and ?/-axes in the desired way. The 
two graphs are shown in Fig. 144. 

Certain properties of the graph of the equation y"^ = x are at 
once evident from the form of the equation : The graph is 
symmetric with respect to the a^axis ; for, if a point (Ji, k) 
satisfies the equation, the point {Ji, — k) also satisfies the 
equation. Why ? The graph lies at the right of the aj-axis ; 
for, any negative value of x would give rise to imaginary 
values, of y. Why? 

- * A table of square roots will facilitate the work. 



X, § 176] IMPLICIT QUADRATIC FUNCTIONS 



267 











I 


i 


i 


t ,i / 3 


:_:_V-- — ^ :: = =— - 




5 , S 1 ^'^^"' 


\ 7-A^ \ 


^/1v_ __ __ _._ _. 


- - 1^ ix^ ::fe^ . ir 2' ■■ ^"^^L 


^s 


4^5; " ~ : J-"--.^ : 


r ._ ..,,, ^f^ "--^ . . 


__±i_J 1 1 1 1 M 1 1 M 1 i 1 11 1 1 1 M 1 m 1 M 



Fig. 144 

The most important properties of the double-valued function 
± Va; to be noted are the following : 

(1) For every positive value of x there are two values of the 
function, viz. -f V^ and — ^x. Therefore the function is two- 
valued. 

(2) As X increases numerically, the corresponding values of 
■\Jx increase numerically, i.e. the numerical value of Va; is an 
increasing function of x. 



176. The Form By'' -f Dx = 0. B^^. 

may always write the equation in the form 

(3) 



Since B^^^ we 



B ' 



Y 


^ 


- 





n>o 


X 



y' 



i.e. in the form 



where n = — D/B. The graph is 
then similar to that of x^^ny, the 
only difference being that the roles 
of the X- and i/-axes are interchanged. If the coefficient n is 
positive, the graph is at the right of the 2/-axis ; if n is nega- 
tive, the graph is at the left of the .v-axis (Fig. 145). In both 
cases the graph is symmetric with respect to the ic-axis, and 



Fig. 145 



268 



MATHEMATICAL ANALYSIS 



[X, § 176 



passes through the origin, at which point it has a vertical tan- 
gent. Why ? The curve defined by an equation of the type 
considered is called a parabola if D =^0. (See Chapter IV.) 
To sketch such a curve rapidly, knowing its general shape, we 
need only plot a few corresponding values of x and y. If i>=0, 
the equation becomes By'^=0. Its graph is then the a^axis. 

177. The Slope of the Curve By^ + Dx = 0. To determine 
the slope of the tangent to the curve 
By^-{- I)x= 0, 

we may proceed by the method used for similar problems in 
Chapters IV and V. To this end we first calculate the change 
ratio Ay /Ax, which is the slope of the chord PQ (Fig. 146). The 



z^ 



\^Au 



TTo 



Fig. 146 
slope of the tangent at P is then the limit which this ratio 
approaches when Ax approaches the value 0. 

Let P(£Ci, 2/i) be any point on the curve, and Q(a;i + Ax, t/i + Ay) 
be another such point. Then we have 

J5(2/i + AyY + i)(a!i + Ax) = 0, 
and 

Expanding the first of these equations, and subtracting the 
second from it, we get 

2 By Ay + BAy'^ + DAx = 0, 



or 



{2By,-rBAy)^^=-D. 



X, § 177] IMPLICIT QUADRATIC FUNCTIONS 269 

Hence, the desired change ratio is 

Aj/_ D 

Aa? 2Byi-\-BAy° 

When Ax approaches zero, Ay also approaches zero. Why? 
The desired slope of the curve 

Bi/ + Dx = 

at the point (xi, y-^ is, therefore, 

^^ 2By, 

The expression for the slope exhibits certain properties of 
the curve : 

(1) The curve has a vertical tangent at the origin (2/1 = 0). 

(2) The slope of the curve above the it'-axis is positive, if B 
and D have opposite signs ; and negative, if B and D have the 
same sign. 

(3) The slope of the curve decreases indefinitely in absolute 
value as the point (a^i, 2/1) recedes indefinitely from the origin. 

EXERCISES 

1. For each of the following equations, determine the slope at the point 
(xi, y\) and sketch the curve represented. For each point plotted deter- 
mine the slope of the tangent and draw the tangent. 

(a) 2/2-4x^:0; (6) ?/2 + 2a; = 0; (c)4x2-3y = 0; 

((?) 4 2/2 + 9x^0; (e) y^ = Qx. 

2. Derive the equation of the tangent to each of the curves in Ex. 1 at 
the point indicated : 

(a)(l,2); (6)(_2,-2); (c)(-3,12); (d) (-4, - 3) ; (e)(6,6)'. 

3. Show that the equation of the tangent to the curve y2 = 2 px at the 
point (xi, yi) on the curve is y\y =p (x + xx). 

4. Draw the curves y'^ = nx for several different values of n on the 
same sheet of paper. It is suggested that the values w = l,2, 5, — 1, — 2, 
be included. 



270 MATHEMATICAL ANALYSIS pC, § 178 

II. THE FORM Ax"^ + By^ + C= 

178. The Case A=^B. The Equation x"" -^ y^ = a\ It so 
happens that, if the units on the x- and ?/-axes are equal, we can 
interpret the left-hand member of this equation geometrically. 
For, it is evident from the figure (Fig. 147) that, under the 




Fig. 147 

hypothesis of equal units, x^ -\- y^ is the square of the distance 
of the point (x, y) from the origin. Hence the equation 

(5) a;2^2/' = «^ 

states that the point {x, y) is distant a units from the origin. 
It follows that the points {x, y) satisfying this equation are all 
on the circle described about as center with the radius a, and 
conversely the coordinates of every point on this circle will 
satisfy the equation. The graph of the equation x^ -\-y'^= a^ is 
then a circle, if the units on the two axes are equal. 

If the units on the axes are unequal, the ordinates of the 
above circle must be shortened or lengthened in a certain ratio, 
according as the unit on the ^/-axis is less than or greater than 
the unit on the a^axis. In either case the graph of the equa- 
tion will be a closed curve. 

Throughout the remainder of this chapter, however, we shall 
assume, in order to fix ideas, that the units on the axes are equal. 

If A = B {AB ^ 0), the equation 

(6) A^-^Bf-^-Q^^ 



X, § 179] IMPLICIT QUADRATIC FUNCTIONS 



271 



may be written in the form x^ -\- y^ = — — » 

The graph of this equation is a circle, if — C/A is positive. If 
— C/A is negative, the equation has no graph, i.e. no pair of 
real values of x and y can satisfy it. If C — 0, tHe only point 
satisfying the equation is the origin.* 

179. The Case A >0, B >0. Consider first the special 
case x"^ + 4t y- = 9. If we solve this equation for ?/, we have 



(7) 



y = ±^^9-x\ 



Now, we know from § 178 that the graph of the function 

(8) 2/ = ±V9^=^ 

is a circle with center at the origin and radius equal to 3. 



Y 




^-^ '~~-^ 




^ \ 




>^ "^i 




' -~ O-liji X 


y> VV 


T "^^ ^^ Z 


. , 


X .^ 


■^^ ^--^ 







Fig. 148 

The ordinates of the points of (7) are then equal to one half 
the corresponding ordinates of the points on the circle (8). The 
construction of the graph of (7) should then be clear from the 
figure (Fig. 148). The graph in question is a closed curve, 
having a greatest length of 6 units and a greatest width of 3 
units. It is symmetric with respect to both axes. 

* The last locus may be considered as a circle with radius equal to 0; it is 
sometimes called a poini circle. 



272 



MATHEMATICAL ANALYSIS 



[X, § 179 



(9) 



The general form 



Ax^-^ By^-j-C=0 



can be treated similarly, if A and B are both positive, 
equation may be written in the form 

C 

A 



The 



(10) 



x^ + ?f = 



This shows that there is no graph if the right-hand member is 
negative. If the right-hand member is 0, the point (0, 0) is the 
only point satisfying the equation. There remains only the 
case where — C/A is positive. 
Equation (10) gives _ 

(11) ,=±^|.^_|_., 

Now, the equation 



(12) 



y 



represents a circle. Equation (11) tells us that the desired 
graph is obtained by shortening or lengthening the ordinates 
of this circle in the ratio ^A/B to 1. 

I >^fTTNL I I I I I I Example. If we solve the equation 9x^ 

7^^1l^^tIIII -\-4y^ = S6 for?/, we obtain y = ± |V4 — x^; 
/[ I mI I I r\ I I I I this tells us that the graph of the given 
equation is obtained from that of the circle 
y = ± V4 — x'^ by lengthening the ordinates 
of the latter to three halves their original 
length. Figure 149 exhibits the result. 



mm 



The graph of an equation of the form 
Fig. 149 Ax^-\-By^-}-C=0 under the hypothesis 

that A and B are both positive and that C is negative, is then 
a closed curve symmetric with respect to both axes. 

The curve represented by an equation of the form (9) above 
is called an ellipse. An ellipse is symmetric with respect to 



X, § 180] IMPLICIT QUADRATIC FUNCTIONS 273 

each of two perpendicular lines, called the axes of the ellipse. 
The intersection of the axes of an ellipse is called the center 
of the ellipse. Knowing the general shape of the curve, the 
quickest way to sketch it from the equation is to find the 
intercepts on the axes and draw a symmetric curve through 
the four points thus obtained. In the example 9 x^ -\- 4: y"^ == 36 
already considered, we find the intercepts to be ic = ± 2 (found 
by placing 2/ = 0) and y = ±o (when x = 0). If we mark the 
four corresponding points, the curve can be sketched readil}. 

EXERCISES 

1. Discuss the locus of each of the following equations and, if the 
equation has a locus, sketch it and show how it is related to a certain 
circle (if the locus is not itself a circle) : 

(a) x^ -\- y^ = 16. (d) ix^-\-y^-\-16 = 0. (g) ix^ -\- Sy^ = 12. 

(c) 4 0:2 4-2/2-16=0. (/) 2x^-{-2y2 = 5. ^ ^ 4 "^ 9 

2. For what values of x in each of the equations in Ex. 1 doesy become 
imaginary ? For what values of y does x become imaginary ? 

3. Show directly from the equations that each of the graphs in Ex. 1, 
if it exists, is symmetric with respect to both the x-axis and the y-axis. 

4. According to the definition above, is a circle an ellipse ? 

180. The Slope of the Curve represented by Ax^ + By^ 
-\- C = 0, Here again we calculate the change ratio Ay /Ax, 
which is the slope of the secant joining the points P(xi, y^) and 
Q{xi + Aa;, y^ + Ay) on the curve, and then find the limit which 
this ratio approaches when Q approaches P along the curve, i.e. 
when Ax and, consequently. Ay approach 0. The calculation is 
as follows : 

Since P and Q both lie on the curve 

, Ax' + By^+C^O, 

we have 

(13) Ax^^-{-By,^+C=0, 

T 



274 MATHEMATICAL ANALYSIS [X, § 180 

and 

(14) A{x, + Axy + B(y, + Ayf +C=0. 

Expanding the squares in the last equation and subtracting 
(13) from (14), we have 

2 Ax^Ax + AAx"^ + 2 By^Ay + BAy^ = 0, 

or (2 By^ + BAy) Ay = — {2 Ax^ + AAx) Ax, 



whence we obtain the slope of the line PQ, 

Ay _ _ 2 Axi -f AAx 
Ax~ 2 By I + BAy 



{B=^0). 



When Ax and Ay both approach 0, we get for the slope of the 
curve at the point (xi, y^) 
(15) m = -^. 

« 
An interesting verification of this result may be noticed. It is well 
known that the tangent to a circle at a point P is perpendicular to the 
radius OP. Now consider a circle with center at the origin. The slope of 
the radius through (xi, yi) is then clearly yi/xi. The slope of the tan- 
gent should, therefore, be —Xi/yi. But this is exactly what the preceding 
formula for the slope gives, when the equation represents a circle, i.e. 
when A = B. 

EXERCISES 

1. Show from the result of the last article that at the points where the 
curve Ax^ + By^ +(7=0 {ABC =^ 0) crosses the ?/-axis its tangents are 
horizontal ; and that at the points where it crosses the a;-axis its tangents 
are vertical. 

2. Find the equation of the tangent to each of the following curves at 
the point indicated. Check the result by sketching the curve carefully 
and drawing the tangent from its equation, 

(a) 4 a:2 4- y2 = 25 at (2, 3). (6) x^ + iy^ = S&t (2, 1). 

(c) 3 x2 + 4 1/2 ::^ 16 at (2, -1). 



X, § 181] IMPLICIT QUADRATIC FUNCTIONS 275 

181. The Case i4 > 0, 5 < 0. We may always write the 
equation (9) so that A is positive. The case where A and B 
have unlike signs leads to a new type of graph. 

The Graph of x^ — ]p- = 9. In seeking the graph of this 
equation, we observe first the following facts : 

(1) The graph crosses the a;-axis at the points (3, 0) and 
(—3, 0), and does not cross the y-axis. Why ? 

(2) The curve is symmetric with respect to both axes. For, 
if the point (/i, li) is on the curve, so also is the point (A, — H). 
Hence, the curve is symmetric with respect to the a>axis. 
Similarly, if the point (/i, fc is on the curve, so also is the 
point (— /i, Iz). Hence the curve is symmetric with respect 
to the 2/-axis. 

(3) Solving the equation for y gives us 



(16) 2/ = ± Va;2 _ 9, 

This incidentally again establishes the symmetry of the curve 
with respect to the x-axis. But it shows further that, if a'2<9, 
y is imaginary. Hence, no part of the curve lies in the strip 
of the plane between the lines x — Z and a; = — 3. In other 
words all values of x between 3 and — 3 are excluded. Solv- 
ing the equation for x gives 

a; = ± V2/' + 9. 
This shows that no values of y are excluded, since 2/^ + 9 is 
positive for every real value of y. 

(4) The slope of the curve at the point (a^i, y-^ is by § 180, 

m =— • 

2/1 
This shows that the curve crosses the a;-axis vertically, i.e. the 

lines a; = 3 and aj = — 3 are tangent to the curve at (3, 0) and 

(— 3, 0) respectively. 

With these results in mind we now calculate the coordinates 

of a few points on the curve and the slope of the curve at these 



276 MATHEMATICAL ANALYSIS 

points. We thus get the following table : 



[X, § 181 



X 


3 


4 


5 


6 


y 





V7 


4 


3V3 


m 


QO 


fvy 


f 


fV3 



We plot these points and those symmetrically situated with 
respect to the two axes and get Fig. 150. We know from 
equation (16) that y increases numerically from as a; increases 



: :: :::::;: : : x::: 


-f- -- 


x ' 


^ ~ G J2 


A-:^ _ ^ TLt- : 


"s \ r, Ay 


"sZs ~ " ^ y " 


_s^s_ __; ^ _ _ <L I 


% \ z ^ I i : 


S ^ 5 /La 


\ ^ ^ ^ ' 


5^1, it 


L S ^ Z J 


\ ^ 1 ^ i 


A si / t 


_ j: sc dt 


1 ^ V i. aL 6 'C 


i zj's Jl 


t ^ ^5 


y -/ - S^ ^L 


t ^ S 5 


J /' S s 


- ^- z : :s^^. 


.-^t.^ s ^^ 


2 2 - - ^s S: """ 


t -,Z_ \i s 


z:? _ _ s s 


^/ - - ^^ 


z_ - . ^ 






i 



Fig. 150 

numerically from 3. We have already seen that the curve 
consists of two branches. It remains only to consider what the 
character of the curve is for numerically large values of x. 

Equation (16) tells us that y increases numerically without 
limit, as x increases indefinitely in absolute value ; i.e. the curve 
recedes indefinitely from both axes. It recedes, however, in a 
very definite way. For, consider the slope m of the curve at 
any point (aji, 2/1). From § 180 we have, for A = l, J5 = — 1, 



X, § 181] IMPLICIT QUADRATIC FUNCTIONS 277 



m=^= ^ 



Vi ± Va^i^ - 9 

the upper sign being used if yi is positive ; the lower, if y^ is 
negative. To fix ideas, let (x^, yi) be a point in the first quad- 
rant and let it move out along the curve indefinitely. We de- 
sire to see what happens to the slope of the curve under this 
condition ; i.e. when Xi becomes indefinitely large. To this end 
we write m in a more convenient form, as follows : 




2/ -2/] =-{^-^i)y 



which shows that as Xi increases indefinitely, m approaches 

more and more nearly the value -|- 1. This shows that the 

further the point {x^, y^) travels out along the curve in the first 

quadrant, the more nearly does the direction of its motion 

make an angle of 45° with the ic-axis. 

Consider now the equation of the tangent to the curve at 

the point {xi, ?/i) : 

2/-2/] = 

2/1 

or, 

^1^ - 2/i2/ = ^i^ - Vi^ 
or. 

This may be written 

xi 9 
y =— ' X 

2/1 2/1 

As Xi and yi become indefinitely large, the slope Xi/yi, as we 
have seen, approaches the value + 1, while the term 9/2/i evi- 
dently approaches the value 0. Therefore, the tangent to the 



278 MATHEMATICAL ANALYSIS [X, § 181 

curve at the point {xi, 2/1) approaches the line 

y=:x. 

A line, which is the limiting position which the tangent to a 
curve approaches, as the point of contact recedes indefinitely 
along an infinite branch of the curve, is called an asymptote of 
(lie curve. 

If the point (x^, y^ recedes indefinitely along the curve in 
the third quadrant {x^ < 0, 2/1 < 0), the slope is positive and the 
tangent approaches the same limiting position as before, 
namely, y = x. Similar considerations (or the symmetry of the 
curve) show that the line 

y = -x] 

is also an asymptote. The two asymptotes are also shown in 
the figure as they are a great help in drawing the curve. 



The Graph of x"^ — y"^ 



If, in place of the 9 in the 



equation x"^ — y'^=^ just considered, we have any other positive 
number, say a^, the discussion is very similar and accordingly we 
can be brief. The curve of the equation x'^—y'^ = a^ crosses the pr- 
axis at the points (a, 0) and (—a, 0), 
and does not cross the 2/-axis. It 
is symmetric with respect to both 
axes. We have y = ± -y/x^ — a^ 
and m = Xi/y^. We find also 
1 




m= ± 



Fig. 151 



4 



from which we conclude that the curve approaches indefinitely 
near the straight lines y—x and y=—x. The curve is, then, as 
drawn in Fig. 151. 



X, § 181] IMPLICIT QUADRATIC FUNCTIONS 



279 



The General Case, when C is Negative. Any equation 
of the form 

where A is positive and B and C are both negative, may now 
be treated without much difficulty. Any such equation can be 
written in the form 

(17) x^ - ny = a\ 

From this we obtain 

n 



a\ 




Fig. 152 



But this shows at once, by com- 
parison with the last equation 
considered, that the ordinates of 
points on the curve x^—n^if-^d?- 
are to the corresponding ordinates of the curve ^ — 'ip- =. o? as 
X/n is to 1. In Fig. 152 we have drawn both the curve 
a;2 — 2/2 == d?- and the curve x"^ — ^y"^ = a^, the ordinates of the 
latter being just one half of the corresponding ordinates of the 
former. The asymptotes of the latter are the lines y^^x and 
y = -^x. 

Since the asymptotes are a great help in sketching the curve, 
we should have a means of obtaining their equations quickly 
from the equation of the curve. From the result of § 180 
(A=l, B=— n^) and considerations similar to those used in 
the discussion of x'^—y^=9, we find the equations m. the 
asymptotes to be 

y =- X and y = x, 

n n 

OT X— ny = and x-\-ny = 0. But these equations are found 
by placing equal to zero each of the factors of the left-hand 
member of the equation of the curve x^ — n^y^ = a^. 



280 



MATHEMATICAL ANALYSIS 



[X, § 181 



An example will show how these various results may be applied in 
sketching a curve whose equation is of the form considered. To sketch 

the graph of 4iX^ — 9y'- = 36, we draw 
first tlie asymptotes 2x—3y=0 and 
2 X + 3 y = (Fig. 153). We next 
place y = 0,in the given equation and 
find the a;-intercepts to be x = 3 and 
£c = — 3. We can now sketch the 
curve with considerable accuracy, since 
we know what its general charac- 
teristics are. 



""■•Jnxrr t ;lL^' " 




: ■ -hi: :: ± :; : _;<?-- : 


S S T , 4^ , - 


4- ^ ^s -( ^^ / - 






:_:--:±:j:___± i^.-zzt'.:.'- : 


:i::::±:±::i^?.-±{3. .± 


J /:^5-_^ \ 


::::::±^i^::.::::-::!s-U:±:i:: 


---^5"^-:-::::-::::::^^--^ 


/''^(^ _ - - ■^s*>. lX 


4Ur<Tl T>T*sj I 



Fig. 153 



The graph of any equation of the form 

^2 _ „2^2 3= ^2 {^n ^ 0, a ^ 0) 

is a curve called a hyperbola. We have seen that it consists 
of two branches ; it is symmetric with respect to each of 
two lines, which are called the axes of the curve. One' of 
these cuts the curve in two points and is called the trans- 
verse axis ; the other axis does not meet the curve at all. The 
intersection of the axes of the curve is called the center of the 
curve. The branches of the curve extend indefinitely and 
approach two straight lines, the asymptotes of the curve, 
which pass through the center. 

We may now complete the discussion of the graph of any 
equation of the form Ax^ -\- By^ -\- C = 0, under the hypothesis 
that A is positive and B negative. We have already disposed 
of the case < 0, by considering the form x^ — 7iy — a,\ The 
case C > leads similarly to the form x^—n^y'^=—a^. By 
interchanging x and y this reduces to the form n^x"^ — y^ = a^ 
which on division by n^ reduces to the case O < already con- 
sidered. The graph of an equation Ax"^ -\- By"^ -\- C=0, when 
A is positive, B negative, and G positive, is therefore a hyper- 
bola with the center at the origin and with its transverse axis 
coinciding with the 2/-axis. 



X, § 182] IMPLICIT QUADRATIC FUNCTIONS 



281 



-^.= ^ =^^^- 


iii^?^-i:-S?"-:i 


- _ -l:^^Z 


^^ ^^12] r 


-P^$ 




'^ T T l^h 



Fig. 154 



It 


a 


^^ ^-^ 


^ » ^'^ 




^v y^ :_ 


_i ^-.^^^ 


' =^'^ ±s. " 




-,«: _ i 





Fig. 155 



The following example will illustrate the method of sketching the 
curve : Sketch the graph of ic^ — 4 ?/2+ 4 = 0. The asymptotes are x — 2 y = 
and X + 2 1/ = (Fig. 154). Placing x = 0, 
we find the ^/-intercepts to be +1 and — 1. 
Having marked the corresponding points and 
drawn the asymptotes the graph is readily 
drawn. 

Finally, when 0=0, the equation 

may be written in the form x^—7i'^y^=0. 

This may be written 

(x — ny){x + 7}y) = 0. This equation will be 
satisfied by all points which satisfy either 
x—ny — or ic-f-7?y = 0, and by no others. 
The locus of the equation is then two straight 
lines passing through the origin. Figure 155 
shows the locus of the equation 4 a;^— 9 2/^=0. 
182. The Case il = or 5 = 0. If ^ = 0, 5 > 0, the equation 
Ax^-{-By^-\-C=0 becomes By^ -\- C — 0. If O is positive, there 
is no graph. If C is negative, the graph consists of two lines 
parallel to the £c-axis. If C is zero, the graph is the x'-axis. 
When B—OjA>0, the graph of the equation consists similarly 
of two straight lines parallel to the t^-axis, if C is negative ; of 
the ?/-axis, if is zero ; and there is no graph, if G is positive. 

EXERCISES 

1. Sketch the graph of each of the following equations : 

(a) x2 - 9 ?/2 = 16. (d) 9 x'^ - 16 ^2 _^ 16 = 0. (g) 3 x2 - 2 y^ = 6. 
(&) x2~9i/2=- 16. (e) 9x^- 16?/2- 16 = 0. (/i) 3x2 - 12 = 0. 
(c) x2 - 9 2/2 = 0. (/) 9 x2 -'l6 1/ = 0. (0 3x2 + 1 = 0. 

2. Give a detailed discussion of the graph of the equation x2— y^ =— 9 
(analogous to the discussion of x2 — y2 _ 9 given in the text). 

3. Give a detailed discussion of the graph of x^—n^y^=—a^. Prove, in 
particular, that the asymptotes of this hyperbola are given by x^—7i'^y^=0. 

4. Prove that no tangent to the curve x^ - y^ = a^ has a slope that lies 
between + 1 and — 1. Prove, in general, that no tangent to the curve 
a;2 _ n22/2 = a2 (a =56 0) has a slope that lies between 1/n and — 1/n. 



282 MATHEMATICAL ANALYSIS [X, § 183 

III. THE FORM Ax^ -\- By'' + Dx -^ Ey + C = 

183. Recapitulation and Extension of Previous Results. 

We have seen in the previous sections of this chapter that an 
equation of one of the forms 

By^ + Da; = 0, 

or ^a;2 -1-52/2 4- 0=0 

represents either 

(a) a parabola, with vertex at the origin and axis coinciding 
with the oj-axis or the 2/-axis ; or, 

(h) an ellipse, with center at the origin and axes coinciding 
with the axes of coordinates ; or 

(c) a hyperbola, with center at the origin and transverse axis 
coinciding with the a;-axis or the y-axis ; or 

(d) two straight lines (which may coincide) ; or 

(e) a single point (the point (0, 0)) ; or 
(/) no locus. 

If we replace xhj x — h and yhjy — k, in any of the above 
forms, we know that the graph of the resulting equation is ob- 
tained from the graph of the original equation by moving the 
latter so that the origin moves to the point {h, k) (the axes re- 
maining parallel to their original positions). 

We may then conclude that an equation of any one of the 
forms 
(18) A{x - hy+ E{y - k) - 0, B{y - ky+ D{x -h) = 0, 

or A(x - hy -\-B(y-ky-\-C = 

represents either 

(a) a parabola with vertex at the point {h, k) and axis coin- 
ciding with the line x — h = or the line y — k = 0; or 

(b) an ellipse with center at the point (h, k) and axes coin- 
ciding with the lines x — h = and y — k = 0; or 



X, § 183] IMPLICIT QUADRATIC FUNCTIONS 



283 



(c) a hyperbola with center at the point (/i, k) and transverse 
axis coinciding with the line a;— /i=0, or the line ?/— A:=0; or 

(fZ) two straight lines (which may coincide), or 

(e) a single point (the point (h, k)) ; or 

(/) no locus. 

Now, any equation of the form 
(19) Ax' -\-Btf + Dx + Ey-\-C=0 

can be put in one of the forms (18) by completing the squares. 
The following examples show how this may be done. 



axis'" ^ 



'^. 



Fig. 156 

Example 1. Discuss and sketch the graph of y2_2y-f2x + 7 = 0. 
This equation may be written in the form 

y2_2y=-2x-7, 
or 

l/2_2y + l=-2x-7 + l, 

i.e. 

It is accordingly a parabola with vertex at ( - 3, 1 ) and axis y = 1. The 
graph is given in Fig. 156. 

Example 2. Discuss and sketch the graph 
of a;2 + y2 _ 4 a; _ 6 y + 9 = 0. 

This equation may be w^ritten in the form 

(x2-4x + 4) + (2/2-6?y + 9) =-9 + 4 + 9, 

or 

(a;-2)2 + (?/-3)2 = 4. 

Therefore the given equation represents a 
circle with center at (2, 3) and radius equal 
to 2. (See Fig. 157.) • Fig. 167 















1, C _ 7 




, i''----''^:- 






Hi" ? ^ ■? 1 ^ 

















284 



MATHEMATICAL ANALYSIS 



[X, § 183 



Example 3. Discuss and sketch the graph of 9x^ + IQy'^ — ISx 
+ 64 2/ - 8 = 0. 

This equation may be written in the form 

9(x2_2x+ )+16(?/2 + 4?/+ )=8, 

or 

9(a;2-2a:+l)4-16(?/2+4?/ + 4) =8+9 + 64=81, 

i,e, 

9(x-l)2+16(?/ + 2)2 = 81. 



- ■ ■ i, ...,.-- 














ffffiMffl™ 






::: i::f :::q:: ::::: 


S 41 - T - ( - 


\ ± iL / - : 


"a ±i 


T "'-rr^ 


S -- It - - . 


3 It : 


± it 





Fig. 158 

Hence this equation represents an ellipse whose center is at (1, — 2) 
and whose axes coincide with the lines x = l,y = — 2. The remainder of 
the discussion is left as an exercise. The graph is given in Fig. 158. 

Example 4, Discuss and sketch the graph 
of 9x2-36x-4?/2 + 24?/ = 36. 

This equation may be written in the form 

9(x - 2)2 _ 4(y - 3)2 = 36, 

which is a hyperbola whose center is at (2, 3) 
(Fig. 159). It is left as an exercise to com- 
plete the discussion and prove that the equa- 
tions of the asymptotes are 3(x — 2) + 
Fig. 159 2(y - 3 ) = and 3(a; - 2) - 2{y - 3) = 0. 



■ ■sry' ■ ■ " ^11-. 


^ - ----- 






5 " v" --A-' '- 


■I /ti 


: ::: : -s:::: ni; : ::: 












if / li 


j/ N \ 


^ j^. .... 








1 


:::. ::. :±.: _:±:::;__: 



EXERCISES 

Discuss and sketch the graph of each of the following equations 



1. a;2 + 42/ + 4 = 0. 

2. x2 + ?/2 + 4 X - 8 y + 1 = 0. 

3. x2 - ?/2 + 2 X = 0. 

4. x2 - 4 X + 2/2 + 2 ?/ + 1 = 0. 

5. x2 + 4 X + 2 2/2 + 4 y + 1 = a. 



6. 9 x2 + 4 2/2 - 36 X - 8 2/ + 4 = 0. 

7. 9x2-42/2- 36x + 82/ = 4. 

8. 2/2 + 22/- 12x- 11 =0. 

9. x2 + 15 2/2 + 4 X + 60 ?/ + 15 = 0. 
10. x2 - 3 2/2 - 2 X - 6 2/ + 7 = 0. 



X, § 184] IMPLICIT QUADRATIC FUNCTIONS 285 

184. The Slope of the Curve Ax^ j^ By'^ -\-Dx -{-Ey + C = 0. 

Let P(aJi, 2/i) ^^^d Q(Xi -|- ^x, y^ + A?/) be any two points on 
the curve. Then 

Ax,^ + By,^ + Dx, + ^2/i + C = 0, 
^(a^i -h Ax-)2 + 5(2/1 + Ay)2+ i>(a;, + Aa;) + E{y, + A^/) + C = 0. 

Expanding the second of these equations and subtracting the 
first from it, we have 

(2 Ax, + A^x + B)^x -I- (2 %i -f- SAt/ 4-^) a?/ = 0. 
Therefore the change ratio, or the slope, of the secant PQ, is 

A^ _ _ 2 Ax^ -\- A^.x + D 
A.i-~ 2 By, -\- B^y -\- e' 

If we let Aa; approach zero. A?/ will approach zero also. Why ? 
Therefore the slope of the curve at any point (Xi, y,) is 

2Ax,-\-D 

m = —^ — • 

2By, + E 

Example. Find the equations of the tangent and the normal to the 
curve x^ + 4y'^ — ix-\-2y — S = 3i,t the point (1, 1). 

Solution : The slope of the tangent at any point (xi, yi) is 

8^1+2 
At the point (1, 1) this slope is |. Therefore the equation of the tangent 
isy— l=i^(x— 1) and the equation of the normal is y — 1 = — 5(x — 1). 

EXERCISES 

1. Find the slope of the tangent to each of the following curves at 
the point specified. 

(a) x2 + 2?/- 3 = at (1, l)j 

(b) x^-\-y^-4 = at (1, V3)^ 

(c) a;2-2 2/2 + 5 = at (1, V3); 

(d) 4x^-\-y^-2x-Sy-lO = at (2, 1). 

2. Find the equation of the tangent to each of the curves of Ex. 1, at 
the point specified. 



286 



MATHEMATICAL ANALYSIS 



[X, § 185 



IV. THE FORM Fxy + Dx + Ey + C = 
185. The Graph of xy = a. The graph of the curve 

xy = a 

is symmetric with respect to the origin ; for, if the coordinates 
Qi, k) satisfy the equation, the coordinates {— h, —7c) also 
satisfy it. Since y = a/x, it is evident that x may assume all 




Fig. 160 



Fig. 161 



values except 0. (See § 36.) As x increases numerically 
without limit, the curve approaches the line y = 0, i.e. y =0 is 
an asymptote. Similarly as y increases without limit, the 
curve approaches the line a? = as an asymptote. It will be 
proved later that the curve is a hyperbola, provided a is not 
equal to zero. If a is positive, the graph is as in Fig. 160. If 
a is negative, the graph is as in Fig. 161. If a is zero, the 
graph consists of the two axes x = and y = 0. 

186. The Graph of Fxy -\- Dx -\- Ey + C = 0. If in the equa- 
tion xy = a we replace x hy x — h and yhjy — k, we know 
that the graph of the resulting equation is obtained from the 
graph of the original equation by moving the latter so that the 
origin moves to the point (^, fc), the axes remaining parallel to 



X, § 187] IMPLICIT QUADRATIC FUNCTIONS 



287 






their original positions. It follows that the equation 

{x — h){y — k) = a{a^O) 

represents a hyperbola whose asymptotes are x = h, y = k. 
If a = 0, the equation represents the two lines x =^ h, y = k. 

Example. Discuss and sketch the 
graph of xy + 4x -{-2y = 1. 
First we write 

(ix±?){y±?) = l. 

Then from inspection we see that the 
given equation may be written in the form 

(x + 2)(2/ + 4)=9. 

That is, the graph is a hyperbola whose 
asymptotes are x = — 2, y = — 4. (See 
Fig. 162.) Fig. 1G2 

187. The Slope of the Curve Fxy + Dx -^ Ey + C = 0. It 

is left as an exercise to show that the slope of the curve 

Fxy-^Dx-{-Ey-\-C=0 
at any point (x^, y^ is 

Fy,A-D 
m = — -p— - • 
Fxi -+- E 



EXERCISES 

1. Discuss and draw the graph of each of the following curves : 
(a) xy = l; (b) xy=-l; (c) xy = 2; (d) xy =- 2; 

2. Discuss and draw the graph of each of the following curves. 

(a) xy + 2x = S; (b) xy + 2 x -\- iy = S; (c) xy - 4x + Sy =2, 

3. Draw the family of curves xy = a, taking several positive and 
several negative values of a. How does xy = 0, compare with these ? 

4. Show that any equation of the form 

^ cx + d 
can be reduced to the form given in § 186. 



288 



MATHEMATICAL ANALYSIS 



[X, § 188 



V. THE GENERAL FORM Ax^ + Fxy + By^ + Dx +Ey -\- C = 

188. The Graph. Methods of drawing the graph of an 
equation in the above form will be illustrated by means of the 
following examples. 

Example 1. Discuss and sketch the graph of 

x^ + 2xy + y'-^ -2x-2=0. 

Solving for y, we have y = — x ±y/'2x -\- 2. All values of x less than 
— 1 must be excluded, for these values make 2 x + 2 negative. Similarly, 
since x=—(y—l)± V — 2?/ + 3, it follows that all values of y greater than 

I must be excluded ; for these values make 
— 2 2/ + 3 negative. The a;-intercepts are 
the roots of the equation a:- — 2 aj — 2 = 0, 
i.e. 1 ± y/S. The ?/-iritercepts are the roots 
of the equation y^—2=0, i.e. ± V2. From 
y —— X ± V2 X -{ 2 it is seen that x may 
start with the value — 1 and increase 
without limit. Similarly from x = — (y—1) 
±V— 2y + S we see that y may start with 
the value f and decrease without limit. 
Using the above data and plotting the 
points 




Fig. 163 



X 


-1 





1 


2 


1±V3 


y 


1 


±V2 


1,-3 


-2±V0 






we obtain the graph in Fig. 163. 

This, problem may be approached from an 
entirely different standpoint. Suppose we let 
y' = ± y/2 x-i-2 and y'f = —x. Plotting these 
curves* (Fig. 164), adding the ordinates of 
y' =±y/2x -\-2 to the ordinates ofy" = — x, 
gives us the desired graph. This may be 
done graphically. We have here a shear of 
y' HZ -t- v/2 x + 2 with respect to the line y" = 



Y 





,CS^ it 


-4^ 


s \l 


\ ^D'^^ 


c ^^ 


lS- 


^^ -J 


-^ ^^ 


^^ Sv 


^^ ^ 


\"^>sS 


v^5^ 


^ ^ 


\ ^ 


3_ 


_r 



Fig. 164 



X. (See § 90.) 



* Observe that the equation ?/' =± v'2 a: + 2 is equivalent to y"^ = 2{x + 1), 



X, § 188] IMPLICIT QUADRATIC FUNCTIONS 



289 



Example 2. Discuss and sketch the 
graph of 

2/2 - 2 xy + 2 x2 — 5 a; + 4 = 0. 
Solving for ?/, we have 

y = x± V— x^ -\- dx — 4:. 

Hence, we merely have to shear the circle 

1/ -±V(x- l)(4-x), 

X- + y2 _ 5 a; + 4 = 0, 
with respect to the line y" = a: in order to 
obtain the desired result. (See Fig. 165.) 
The complete discussion is left as an exercise. 
Example 3. Discuss and sketch the graph of 

7 a-2 + 36 xy -S6y^-25 = 
Solving for y, we have 

y = lx± iVl6a;2-25, 







Y. ^'^N -,' 


-X ^'^ 


Z -.2 


y Z i 


-.^-Z I 


t-7^ J^ 


^^v 


y^ Z^^ 


z2^^ \ 


-/ - _ __ _ 


/O X 


^ \ J 


^^^--^ 







Fig. 165 





T7- 


1 




I 


J 


/ ^ 


-^ - ^ " 


U A y> . 


^> - t ,^^%r^ 


^>s ^%^^-r^'i^ 


^ N / X '^^^ i-C 


^ SZ ^ !&^ 


::::::_::n^^::::^^:^: :_:_:::: 


-::^:::^--:^^-:S-J4^^^p- 




~ ) .^^"^ ^^'■' , *--U. 


_ 2ES'' 1 N,^. 


<^ N 5^^». 


yT^^i ^^l ' 


^^^Jl^\r •' ^ss 


^^ ^ - t ^5^ 


^ J Si 






?' ' it 


^z : 


^2 - 






1 



Fig. !()(> 
which shows that the desired graph may be obtained by shearing 



I.e. 



y=±iVl6x2-25, 
16a;2_36i/2_25 = 0, 



with respect to the line y = \x (See Fig. 166.) The complete discussion 
is left as an exercise. 



290 MATHEMATICAL ANALYSIS [X, § 188 

EXERCISES 

Discuss and sketch the graph of each of the following equations : 

1. Ax"^ + y"^ — i xy - X -\- S z= 0. 4. iy^ - ixy + x:^ = 1 - x. 

2. 7f-2xy + 3x = 2. 5. Qy'^ -12xy -\-'Sx^ + ox = 6. 

3. 2/2-8 xy +iex'^=l- x2. 6. y^ - 6xy -\- Sx^ - \0x - 25 = 0. 

189. The Slope of the Curve Ax^ + By"^ + Fxy -\- Dx + Ey 

+ C = 0. It is left as an exercise to prove that the slope m at 
any point {xi, y^) is 

. 2By,-^Fx,+E 

EXERCISES 

Find the equations of the tangent and the normal to each of the follow- 
ing curves at the points indicated. 

1. 48 x2 - 11 a;y - 17 2/2 - 129 a: + 24 ?/ + 81 = ; (2, 1), (3, - 3). 

2. x?/ 4- 2 X - a;2 4- ?/2 + 6 2/ = ; (0, 0), (0, - 6). 

3. 81 y2 + 72 xy + 16 x2 - 96 x = 378 y - lU ; (3, 2). 

190. A General Theorem. The results of the examples and exercises of 
§ 188 suggest that the graphs of equations of the second degree involving an 
xy-term are similar to the graphs of equations of the second degree in which 
the xy-term is lacking. We may now prove that this is a fact. The 
theorem is as follows : 

Any equation of the form Ax^ + Fxy + By- + Dx -{■ Ey + C = repre- 
sents either an ellipse, or a hyperbola, or a parabola, or two straight lines 
{which may coincide), or a single point, or no locus. 

We shall prove this theorem by showing that if the locus of the 
equation 

(20) Jx2 + Fxy -{- By"^ + Dx + Ey -\- C=0 

be rotated about the origin through sl properly chosen angle d, its equation 
will be of the form 

(21) ^'x2 + Bhf + D'x + ^'y + C = 0. 

The theorem then follows from § 183. 



X, § 190] IMPLICIT QUADRATIC FUNCTIONS 291 

We saw in § 137 that, if any point P(x, y) be rotated about the origin 
through an angle ^ to a new position P'(a;', y'), the coordinates of P and 
P' are connected by the relations : 

x' =x cos d — y sin 6, 
xsin d -\- y cos d. 



(22) 



Solving these equations for x and y in terms of x', y\ we obtain 
X = x' cos d -\-y' sin d, 
^ -^ y =— x' sin 6 -\- y' cos 6. 

If P(x, y) satisfies equation (20), P'(x', y') will satisfy the equation ob- 
tained by substituting the values of x, y from (23) in equation (20) . 
The result of this substitution is as follows : 

A (x' cos d -\- y' sin d)^+ F(x' cos^ + y'sin e){—x' sin d -{- y' cos 6) 
+ B(— x' sin e + y' cos ey 
+ D{x' cos d + y' sin d) 
+ E(—x' sin e + J/' cos ^) + O = 0. 

When expanded and rearranged according to the terms in x', y\ we 
obtain 

(24) A'x'-^ + F'x'y' + B'yi'^ + D'x' + i^'?/' + C = 0, 

where A' = A cos^ + P sin2 ^ - P sin ^ cos 6. 

F' = 2(A- B) sin ^ cos ^ + P(cos2 6 - sin^ ^). 
B' = A sin2 ^ + P cos2 ^ + P sin 6 cos ^. 
D' = Bcos e— E sin ^. 
E' = D sin ^ + P cos ^. 
C" = C. 

Equation (24) will be of the desired form (21), if the angle 6 is so chosen 
that F' = 0. Now, F may be written 

(25) P' = (^-P)sin2^ + Pcos2^: 

F will, therefore, be equal to zero, if 

tan 2 ^ ^ 



B-A 



A value of 6 satisfying the condition (26) can then always be found.* 
This completes the proof of the theorem. 

The following exercises will illustrate the above proof. The method 
may also be used to draw the graphs of equations involving the xy-term. 

* If ^ = ^, we take 26 = 90°, i.e. d = 45°. 



292 



MATHEMATICAL ANALYSIS 



[X, § 190 



EXERCISES 

Determine the angle ^through which the loci of the following equations 
must be rotated in order that their new equations shall contain no xy-term. 
Determine the new equation and use it to draw the locus of the original. 

1. Sx^ + 4xy -\-5y'^-36 = 0. 

Solution : After substituting x =x' cos -\- y' sin ^, y z=— x' sin d 
4- y' cos 6, the equation becomes 
(1) (8 cos2 ^4-5 sin2 6- 4: sin 6 cos e)x'^ 

+ [6 sin d cos e-i- 4(cos2 d — sin^ e)'\x'y' 
+ (8 sin2 6+ b cos^ ^ + 4 sin ^ cos 0)i/'2 _ 35 _ q. 
2 tan e 



Therefore, 



tan 2 ^ = — 



3 1 - tan2 d 

Solving this equation for tan 6, we have 

4 tan2 ^ _ 6 tan — 4 = 0, 
or tan ^ = 2 or — ^. 

We choose tan 6 
rant) ; therefore 
2 

V5' 



2 (^ in first quad- 















. 








5 ' 




. "T^ ^ 




._ ^ - 


^s ^ 


/ \" 


-S ^"^ 


t ^' 


S.^J' 


jj ^ 


^^ 


ji:_.i 








\^ 


\ 1 / 


^'\ 


\ J^ 


^^ A 


^ mrcltan 2 








"■ \ y' 




\ 
















. 



Sin 



cos d . 



1 



Substitutin, 
obtain 



2. x2 - ?/2 _|_ 2 xy - 12 = 0. 



5 these values in (1) we 

4 x'2 + 9 ?/'2 = 36. 

The desired graph is obtained from the 
grajjh of this equation by rotating it 
through the angle — Q about the origin. 
The construction of the adjacent figure 
explains itself. 

5. 3 x2 - 2 a;?/ + 1/2 ^ 6 = 0. 



12 = 0. 



6. 8x2 



12 xy + 3 y2 _ 36 = 0. 

3 ?/2 + 42 = 0. 



3. x2-y2_|.2x?/ + 2x 

4. xy = 4. 7. 2 x2 - 12 xy 

8. 6 x2 + 4 xy - ?/2 + 48 X - 12 y - 10 = 0. 

9. 9 2/2 + a;2 + 2 xy = 0. 

10. Prove that the locus ot xy = c may be rotated about the origin so 
as to coincide with the locus of x^—y'^ — a^, provided a^ =±2c. 

11. With the notation of § 190, prove that A' -\- B' = A -\- B and that 
{A' - B')^ + F'-^ =(A- BY + F\ 



PART III. APPLICATIONS TO GEOMETRY 

CHAPTER XI 

THE STRAIGHT LINE 

191. Introduction. We have hitherto used coordinates pri- 
marily for the purpose of representing functions graphically 
and investigating the properties of those functions. We have 
seen that every continuous function defines a curve or a 
straight line, the graph of the function. Thus far, we have 
laid emphasis only on the discovery of the characteristics of the 
functions from the known properties of the curves that repre- 
sent them. 

Conversely, we have seen that every curve or straight line, 
in the plane of a system of rectangular coordinates, defines a 
function ; i.e. the points of any such curve associate with every 
value of X one or more values of y. If this function can be 
determined when the curve is given, the properties of the 
curve may be studied from the properties of the function. 
This function is usually expressed by means of an equation in 
X and y, called the equation of the curve. We propose now to 
study the properties of various curves by means of their equa- 
tions. (See § 62.) 

Up to this time, we have used different scales on the two 
axes whenever it was convenient to do so. Throughout this and 
the next four chapters we shall assume, unless the contrary is 
specifically stated^ that the units on the x- and y-axes are equal. 



294 



MATHEMATICAL ANALYSIS [XI, § 192 



192. The Distance between two Points. Given the two 
points Pi {xi, yi) and P2 {x2, y^)^ let us find the length of the 
segment PiP^- If a line be drawn through Pj parallel to the 
ic-axis and another through P2 parallel to the 2/-axis to form 
the right triangle P1QP2 (Fig. 167), we have at once 



(1) 



P,P,=^J\Q' + QPl 



T 






P. 




^ 




N, 


Pr^ 




Q 









Mr 


M,'X 










Fig. 



Pi 



M, 



N, 



JU 



The segment PiQ is equal to the projection M^M^ of PxPi on 
the a^-axis and (^P^ is equal to the projection NiN^ of P1P2 on 
the 2/-axis. By the result of § 37, we have 

PiQ= M^M^ = x.^— ccj, 

QA=i^==2/2-2/i. 
Substituting these values in (1), we have the desired formula : 
(2) 



PxP,=V(x2-Xiy-j-{y,-yi)\ 



193. The Simple Ratio. Given two distinct points Pj, P2 
and any point P (distinct from P2) on the line P1P2, the ratio 
P1P/PP2 is called the simple ratio of P with respect to Pi, P2. 

The line-segments in this definition are directed segments. 
Accordingly the simple ratio of P with respect to Pi, P2 is 
positive if P is between Pi and P2, and negative if P is on 
either prolongation of the segment P1P2. 



XI, § 194] 



THE STRAIGHT LINE 



295 



194. Point of Division. The coordinates (x, y) of the point 
P on the line joining Pi (x^, y^ to P2{x2y 2/2) such that the 
simple ratio 






are given by the formulas 
(3) 



;C=£l±J^, ^^J/l + Xyg. 



1 + X 



l + X 



Proof. Draw lines through Pi, Pg, 
P parallel to the axes, meeting the 
jc-axis in Mi, M^, M, and the y-axis 
in Ni, N2, N, respectively (Fig. 168). 
Then, since P1P/PP2 = X, we have 

MiM ^ ^^=X 

MM^ ' NN, 



The first of these relations gives (by § 37) 

X2 — X 

Solving this equation for x gives 

Xi + \X2 



r 

N 
Ml / 




P 


M 


■^2 


/ 




X 


Pi N, 


Q 



Fig. 168 



x = 



1+A 



Similarly from the second relation above we obtain 



y = 



_ 2/1 + ^2/2 
1+X 



The mid-point of P1P2 is obtained from the value A = 1. Why ? 
Accordingly the coordinates of the midpoint of P1P2 are 



f xi + X2 Vi + y2 \ 
l~2-' 2 ) 



296 MATHEMATICAL ANALYSIS [XI, § 194 

EXERCISES 

1. Find the distance between the following pairs of points : (1, 2) 
and (5,3); (- 1, 6) and (2, -3); (-2, -1) and (-1,4); (-3,4). 
and (1,4). 

2. Find the lengths of the sides of the triangle whose vertices are 
(—1, 1), (4, — 4), and (1, 3). Prove that it is a right triangle. 

[Hint : A right triangle is the only kind of triangle in which the square 
of one side is equal to the sum of the squares of the other two sides.] 

3. An isosceles triangle has its vertex at (4, 4) and the vertex of one 
of its base angles at (0, — 1). The vertex of the other base angle is on 
the a;-axis. Find the coordinates of the latter vertex. 

[Hint : Let the unknown point be (ic, 0) and equate the equal sides. 
How many solutions are there ?] 

4. Find the coordinates of the point whose simple ratio with respect 
to (2, 1) and (—4, 7) is 2. Find the coordinates of another point 
whose simple ratio with respect to the same two given points is — 2. 

Draw a figure illustrating this problem. 

5. Check the result of Ex. 4 by calculating the lengths of the seg- 
ments involved. 

6. Find the coordinates of the point which divides the segment from 
(2, — 1) to (— 4, 3) internally in the ratio 1 : 4. 

7. Find the coordinates of the mid-points of the sides of the triangle 
in Ex. 2. 

8. A quadrilateral has its vertices at the points (—2, 1), (3, 1), 
(5, 3), and (0, 3). Show that its diagonals bisect each other. What 
kind of a quadrilateral is it ? 

9. Find the coordinates of the points of trisection of the segment 
from (3, -6) to (0, 3). 

10. A triangle has its vertices at the the points (0, 4), (2, — 6), ( - 2, 
— 2) . Find the coordinates of the points two thirds of the way from 
each vertex to the middle point of the opposite side, and thus show that 
the three medians of the triangle all pass through the same point. 

11. The vertices of a triangle are (xi, yi), {xi,y-2), (scs, yi). Find 
the coordinates of the point of intersection of the medians. 

12. Show that the triangle ^1(4, 1), i?(l, 4), C(5, 6) is isosceles. 



XI, § 194] THE STRAIGHT LINE 297 

13. One end of a line whose length is 13 units is at the point (3, 8). 
The ordinate of the other end is 8. What is its abcsissa ? 

14. The middle point of a line is (2, 3) and one end of the line is at 
the point (4, 7). What are the coordinates of the other end ? 

15. The points (2, 1), (3, 4), (— 1, 7) are the mid-points of the sides 
of a triangle. Find the coordinates of the vertices. 

16. Find the area of the isosceles triangle whose vertices are (4, 1), 
(1, 4), (5, 5) by finding the length of the base and the altitude. 

17. What equation must be satisfied if the points (a:, y), (2, 1), (1, 4) 
form an isosceles triangle the equal sides of which meet in (ac, y)? 

18. Prove that the points (- 2, - 1), (1, 0), (4, 3) and (1, 2) are the 
vertices of a parallelogram. 

19. The line from (xi, yi) to (^2, ?/2) is divided into 5 equal parts. 
Find the coordinates of the points of division. 

20. A point is equidistant from the points (2, 1) and (—2, 1) and 7 
units distant from the origin. Find its coordinates. 

QUESTIONS FOR DISCUSSION 

1. Does the distance between two points depend on the order in 
which the points are taken ? Does the formula for the distance give the 
same result no matter in which order the points are taken ? Why ? 

2. Does the simple ratio of a point with respect to Pi, P2 depend on 
the order in which the points Pi, P2 are taken ? What is the relation 
between the simple ratio of P with respect to Pi, P2 and the simple ratio 
of P with respect to P2, Pi ? 

[ Hint. The answer to this question follows most easily from the defini- 
tion of simple ratio. Prove the relation in question by means of the 
formulas in § 194. ] 

3. Can the simple ratio of a point P with respect to Pi, P2 be — 1 ? 
Why ? As the simple ratio approaches —1 what is the motion of P? 

4. What can be said of the position of the point P, if its simple ratio 
with respect to Pi, P2 is positive ? if its simple ratio lies between and 
— 1 ? if its simple ratio is less than — 1 ? 

6. If the simple ratio of P with respect to Pi, P2 is X, what is the 
simple ratio of Pi with respect to P and P2 ? of P2 with respect to Pi 
and P ? 



298 



MATHEMATICAL ANALYSIS 



[XI, § 195 



195. The Area of a Triangle. One Vertex at the Origin. 

Let us try to find the area of a triangle whose vertices are 
0(0, 0), Pi(xi, yi), and P2(^'2j 2/2)- I^^t the angles XOPi and 
XOP2 be denoted by Oi and O2, respectively, and let the angle 
F1OP2 of the triangle have the absolute measure 6 (Fig. 169). 





Fig. 169 

The area of the triangle is then equal to ^ OPi • OP2 sin 0. 
Now, the directed angle P1OP2 differs from 62 — ^1, if at all, 
only by multiples of 360° (§ 101). Therefore 

sin ^ = ± sin (PiOP2)= ± sin {$2 - ^1). 

The area of the triangle OP1P2 is, then, 

A=:±^OPi' OP2 sin(^2 - ^1) 

= ± I OPi ' OPsCsin $2 cos $1 — cos $2 sin ^1) 

(§138) 



= ±iOP,-OP, 



2/2 _^i ^ Jh\ 

OP2 OPi OP2 OPj 



The area of the triangle OP1P2, in the ordinary sense of the 
term, is therefore equal to the absolute value of the expression 

l(xiy2 - XiVi). 

For some purposes it is convenient, however, to regard the 
area enclosed by a curve as a signed quantity, just as we have 



XI, § 195] THE STRAIGHT LINE 299 

found it convenient to regard line-segments and angles as 
signed quantities. 

To this end, we observe that a point moving on the boundary 
of an area may make the circuit in either of two opposite 
directions (Fig. 170). With each of these directions is asso- 
ciated a definite rotation about a ^...,,^^ ^^^ 
point within the area. If the bound- ( ^^X ^ J^^ 
ary is traversed in a direction which \S^ X^x"^ 

produces a positive rotation about a positive arcuU Negative circuit 

point within the area, the circuit and ^^^- ^"^^ 

the area are regarded as positive ; if the boundary is traversed 
in the opposite direction, the circuit and the area are regarded 
as negative. Hence if an area is represented by a signed 
number, the sign of this number tells us the direction in which 
the boundary is traversed. 

In case of a triangle OP1P2 (Fig. 169) the order in which 
the vertices are written determines a direction of traversing 
the boundary. If OP1P2 is positive, OP2P1 is negative, and 
vice versa. Now in going around the triangle in the direction 
OP1P2, a segment OP joining to a point P moving on P1P2 
generates a directed angle PiOP^- This angle is positive or 
negative according as the circuit OP1P2 is positive or negative. 
Moreover the measure of the angle PiOPo differs from 62 — ^1, 
if at all, only by multiples of 360°. The expression 

i OP, ' OP; sin((92 - ^1) 

is, therefore, positive or negative according as the circuit OP1P2 
is positive or negative. We have then finally : 

The area of a triangle OP1P2 is given in magnitude and in 
sign by the formula 
(4) Area OP^P. = lixiUo - ^^2^1)- 



300 



MATHEMATICAL ANALYSIS 



[XI, § 196 



196. The Area of Any Triangle. The convention as to the 
sign of an area is serviceable in deriving a formula for the area 
of any triangle in terms of the coordinates of its vertices. 
Let the vertices be Pi(xi, yi), P2(^) 2/2)) ^3(^*3? Ik)- Join these 
vertices to the origin by lines OPi, OP2, OPs. We novr con- 
sider the three possible cases, according as the origin is inside 



><a! 



^^ 



Fig. 171 




'1^ 


Y 


X- 


\y 





X 


k 







Fig. 172 



Fig. 173 



(Fig. 171), outside (Fig. 172), or on, a side (Fig. 173) of the 
triangle PJ^JP^. Then in all cases, we have 

A P,P,P, = AOP.P, + AOP2P3 + AOP.Pi, 

if due regard is taken of the signs of the areas. Hence 

(5) Area of A P.P^P^^ \{if',x, - x^y, + y,x^ - x^y^ + y^x^ - x^y,). 

It might appear that this formula is difficult to apply. The following 
method makes it very simple. Write the coordinates of the 
vertices in two vertical columns as indicated, repeating the 
coordinates of the first vertex. Multiply each x by the y in 
the next row and add the products. This gives iCi?/2+X2y3+X3i/i. 
Then multiply each y by the x in the next row and add the 
products. This gives yiX2 + y^Xs + ysXi. Subtract the second sum from 
the first and divide the result by 2. The final result will be the area 
sought, with its proper sign-* A similar method may be used for finding 
the area of any convex polygon whose vertices are given. See Exs. 
6, 7, 8, pp. 301, 302. 

* The student familiar with the elements of the theory of determinants 
will .observe that the area can be expressed as 



A=i 



Xi 


yi 


X2 


Vi 


X3 


ya 


Xi 


y\ 



Xl 


V\ 


1 


Z2 


2/2 


1 


Xg 


Vs 


1 



XI, § 197] THE STRAIGHT LINE 301 

Example. Find the area of the triangle whose vertices are Pi(— 4, 3), 
P2(— 1, - 2), PaC— 3, - 1). We write the coordinates of the 
vertices in two columns, repeating those of the first vertex. 
Performing the first step described in the previous paragraph ~ q ~ i 
we obtain 8 + 1 — 9 = 0; the second step gives — 8 + 6 + 4 
= 7 ; the third step gives 0— 7 =— 7 ; dividing this by 2, 
we obtain — 3^ as the area of triangle P1P2P3. The magnitude of the 
area is 3^ square units, and the direction Pi to P2 to P3 is negative. 
Draw the figure and verify the latter statement. 

197. Condition for CoUinearity of three Points. If three 
points Pi, P2, P3 are collinear, the area of the triangle formed 
by them is zero ; conversely, if the area of a triangle is zero, 
the three vertices are collinear. Therefore, a necessary and 
sufficient condition that three points be collinear, is that the right 
hand member of (5), p. 300 be equal to zero. 



EXERCISES 

1. Find the areas of the following triangles and interpret the sign of 
the result in each case. Illustrate by appropriate figures. 

(a) (1, 3), (4, 2), (2, 5). (c) (- 5, 2), (- 4, - 3), (1, -1). 

(6) (2, 4), (- 3, 1), (1, - 7). (d) (a, a), (- 6, - 6,) (c, d). 

2. Show that the following sets of three points are collinear : 

(a) (0,1), (2,5), (-1, -1). (c) (1,-2), (6, 1), (-4, -5). 

(&) (2, 1), (- 4, 4), (4, 0). (d) (0, -6), (1, a-6), (a, a^-b). 

3. The point (h, h) is collinear with (2, 5) and (5, - 3). Find its 
coordinates. 

4. Find the point on the y-axis collinear with (2, 5) and (5, — 3). 

6. Under what conditions on a, b, c, and d are the points in Ex. 1 (d) 
collinear ? Interpret each of the conditions geometrically. 

6. Area of any polygon. Show that the method of § 196 may be ex- 
tended to derive a formula for the area of any polygon in which two 
sides do not cross each other, and that if P1P2P3 -" Pn are the vertices of 
the polygon taken in order around the polygon, we have 

Area of polygon = A OP1P2 + A OP2P3 + A OP3P4 + ... + A OP„Pi, 
if due regard is paid to signs. 



302 MATHEMATICAL ANALYSIS [XI, § 197 

7. Find the area of the quadrilateral whose vertices are (1, 2), 
(-2,3), (-3, -4), and (4, -5). 

8. Find the area of the polygon whose vertices are (4, 1), (2, 3), 

(0,4), (-2,3), (-4,1). 

9. Prove that the points (1, 2), (3, 6), (—1, — 2) are collinear. 

10. Show that the area of the triangle whose vertices are (2, 6), 
(— 4, 3), (—2, 7) is four times the area of the triangle formed by join- 
ing the middle points of the sides. 

198. Applications to the Proof of Geometric Theorems. 

We shall now give a few elementary examples to show how 
the methods hitherto developed may be used in the proof of 
geometric theorems. 

Example 1. Prove that the line joining the vertex of any right tri- 
angle to the mid-point of the hypotenuse is equal to half the hypotenuse. 
Let ABC be any right triangle. In order to apply the methods of 
coordinates we must first locate a pair of coordinate axes. Any two 
perpendicular lines will serve the purpose, but the 
work incident to the solution of many problems 
may usually be greatly simplified if we choose 
the axes judiciously. In this case it is convenient 
(a,o')~x ^° choose the legs of the triangle as axes. The 
F 174. coordinates of the vertices are then (Fig. 174) 

(0, 0), (a, 0), and (0, &). The midpoint of the 
hypotenuse is (§ 194) (a/2, b/2). The length of the line joining this 
point to (0, 0) is V(a72JHjbj2y = ^VoM^ But the length of the 
hypotenuse is Va'^ + b'^. This proves the theorem. 

Example 2. Prove that the diagonals of a parallelogram bisect each 
other. 

Let ABCD be any parallelogram. Let a side of the parallelogram lie 
on the X-axis a vertex being at the origin. 




(See Fig. 176.) We may assign the coordinates 
{a, 0) to the vertex B, and (6, c) to the vertex 
D. The coordinates of C will then be (a-f-&, c). —q 
Why? 

We now calculate the coordinates of the 



•PCb.c) C(a+6.o) 



^ 



£{0,0-) X 
Fig. 175 



mid-point of ^C and also of the mid-point of BD, by the formula of 
§ 194. It will then be seen that the midpoints coincide. 



XI, § 198] THE STRAIGHT LINE 303 

Example 3. Prove that if the lines joining two of the vertices of 
a triangle to the mid-points of the opposite sides are equal, the triangle 
is isosceles. 

Let ABC be the triangle, M^ N the mid-points of 
the sides AC^ BC, respectively, with AN'= BM. Let 
the a;-axis lie along the side AB and let the y-axis 
pass through the vertex C (Fig. 176). Let the coordi- 
nates of A, B, C he (a, 0), (&, 0), (0, c) respectively.* 

We must first state the hypothesis of the theorem 
analytically, i.e. in terms of the coordinates. To this ^^' 

end we note that the mid-point of AC is M = {a/2, c/2), and that 

BM^' 



Y 

/ 


CCo.c) 


J 


> 


P 


^ X 


A(a,o) 


. B(b,o) 



Similarly, we have AN^ = (a-^Y-h- 



By hypothesis, AN = BM. Hence we have 



■ This condition gives 

(-f)=.(«-|> 

which, when simplified, gives either a = b or a=— b. The first result 
would imply that the points A and B coincide, which is contrary to the 
hypothesis, and is therefore rejected. The second result yields readily 
that AC= BC, which was to be proved. 

EXERCISES 

1. Prove analytically that the diagonals of a rectangle are equal. 

2. Prove analytically that the line joining the mid-points of two sides 
of a triangle is half the third side. 

* In the figure a is a negative number. However, the discussion that follows 
applies at the outset to any numbers, a, b, c. It will appear later in the dis- 
cussion that, under the hypothesis of the theorem, a and b must have opposite 
signs. One of the advantages of the analytic method is the fact that it is 
general, and that ordinarily special cases do not have to be considered 
separately. 



304 MATHEMATICAL ANALYSIS [XI, § 198 

3. Prove analytically that two triangles with the same base and 
equal altitudes have the same area. 

4. ABCD is a parallelogram, with A^ C as opposite vertices. JIf and 
iVare the mid-points of the sides AB and CD respectively. Prove ana- 
lytically that the lines AN and CM trisect the diagonal BD. 

5. If P is any point in the plane of a rectangle, prove analytically 
that the sum of the squares of the distances from P to two opposite 
vertices of the rectangle is equal to the sum of the squares of the dis- 
tances from P to the other two vertices. 

6. Prove analytically that, if the diagonals of a parallelogram are 
equal, the figure is a rectangle. 

7. Prove analytically that the two straight lines which join the 
mid-points of the opposite sides of a quadrilateral bisect each other. 

8. Show analytically that the figure formed by joining the middle 
points of the sides of any quadrilateral is a parallelogram. 

9. If ilf is the mid-point of the side BC of any triangle ABC, prove 
that AB2+ AC^ = 2(AM^ + MC'^). 

10. Prove analytically that the distance between the middle points of 
the non-parallel sides of a trapezoid is equal to half the sum of the 
parallel sides, 

11. The difference of the squares of any two sides of a triangle is equal 
to the difference of the squares of their projections on the third side. 

12. Prove that the sum of the squares of the sides of any quadrilateral 
is equal to the sum of the squares of the diagonals plus four times the 
square of the distance between the middle points of the diagonals. 

13. If A^ jB, O, Z) are four points of a line prove the relation (due to 
Euler) : AB ■ CD-\-AC ■ DB-\-AD • BC=0. (The segments are directed.) 

14. If M and iV, respectively, are the mid-points of two segments J.B and 
CD on the same line, show that 2 il/iV= AC + BD= AD + BC. 

15. If 31 is the mid-point of AB and P any other point of the line AB, 
show that PAPB= PM^ - MAK 

16. Two sources of light of intensity a and /3 are situated at the points 
A and B respectively of a line. Find the position of a point on the line 
which is lighted with the same intensity by the two points. How many 
points satisfy the relation ? 

[Hint : The intensity of light at a point varies inversely as the square 
of the distance of the point from the source of light and directly as the 
intensity of the source.] 



XI, § 198] THE STRAIGHT LINE 305 

17. Two objects of weights Wi, w^ are situated at the points A^ Ao. 
The center of gravity of the two objects is defined to be the point of the 
line A1A2, whose simple ratio with respect to Ai, A2 is W2/W1. If Ai, A2 
are on the x-axis, and their coordinates are Xi, X2^ find* the coordinate of 
the center of gravity. Show that the center of gravity does not exist, if 
Wi = — W2. Give an interpretation to a negative w. 

18. Given n weights lOi, wo, ■•■■,Wn situated at the points Ai, A2, -••, A^ 
on a line. Find the center of gravity of ^1, A2 with weights Wi, W2 ; 
then the center of gravity of the point found taken with th^ weight 
Wi 4- W2 and A3 with the weight W3 ; then the center of gravity of this 
new point taken with "the weight Wi + W2 + 103 and A^ with the weight 104 ; 
and so on. Show that when all the ?i points have been used, there is 
obtained a point which is independent of the order in which the points 
were taken. The point thus determined is called the center of gravity of 
the n points. When does no center of gravity exist ? Under what coy- 
ditions is it indeterminate ? Show that if the latter conditions hold, each 
of the given points is the center of gravity of the remaining ones each 
taken with the weight assigned to it. 

19. The first (or static) moment of a point P of weight w about a line 
I is defined to be the product of lo by the distance of P from I. Given n 
points Pi =CXi, yi)(i= 1, 2, .-., n) in a plane with weights lOc, respec- 
tively, determine the coordinates of a point P of weight lOi + «?2+ -•• 4- Wn 
such that its moment about the x-axis shall be equal to the sum of the 
moments about the x-axis of the points P,- and such that its moment 
about the y-axis shall be the sum of the moments about the y-axis of the 
points Pi. The point P is the center of gravity of the set of points. Com- 
pare with the result of Ex. 18. 

20. The second moment or the moment of inertia of *a point P with 
respect to a line I is defined to be the product of the weight 10 of P by the 
square of its distance from the line. Given n points Pi in a plane whose 
distances from a fixed line I are Xi, and whose weights are Wi respectively. 
Let Ml be the sum of the first moments, M2 the sum of the second 
moments of these points about the line I. Let V be a second line, paral- 
lel to the first and h units from it (to the right or left according as h is 
positive or negative), and let Mi and iHf ^ be the sum of the first and sec- 
ond moments of the given points about I'. Let W be the sum of 
the weights Wi -{- 102 + ••• + Wn. Show that 

M'i = Mi-hW and M'2 = Mif-2hMi + h^W. 




306 MATHEMATICAL ANALYSIS PCI, § 199 

199. Directed Lines and Angles. An angle from a directed 
line li to a directed line I2 is an angle through which li must 
be rotated to . make its direction coincide with that of Zg. 

Any such angle we denote by {li y. Clearly 
if li and Zg intersect in a point M (Fig. 177), 
(Zi Z2) is the directed angle from Zi to l^ as 
^^' ' defined in § 98 since the directions of Zj and Z2 

define uniquely the half-lines issuing from M. As we observed 

in § 101, an angle (Zi I2) may have various determinations 

diifering from each other by multiples of 360°. 

The angle from the aj-axis to a directed line Z is called the 

inclination of Z (Fig. 178). If the inclination of a directed line 

Zi'is 61 and the inclination of a directed line l^ 

is $2, the angle from Zi to Z2 is given (§ 101) by 

the equation 

(6) (/i/2)=e2-ei, 

where the equality sign means equal except 

possibly for multiples of 360°. Fig. 178 

200. Undirected Lines and Angles. If two lines Zi and Z2 
are not directed, an angle from Zi to Z2, defined as an angle 
through which l^ must be rotated to make it parallel to Zg, will 

have various determinations which differ by 
multiples of 180° (Fig. 179). The smallest 
positive (or zero) angle from Zi to Z2 is then 
Fig. 179 unique and less than 180°. The inclination of 

an undirected line is defined as the smallest positive (or zero) 
angle through which it is necessary to rotate the a;-axis in order 
to make it parallel to the line. In Chapter III we used the 
slope m of a line to measure its inclination. It follows almost 
immediately from the definition of slope m and inclination 6 
that we have m = tan 0. 





XI, § 201] THE STRAIGHT LINE 307 

To calculate the angle from a line l^ to a line l^ we make use 
of (6), § 199, if the inclinations 0^, 6^ of l^, I2 are known. If 
the slopes mj, m^ of Zj, I2 are given, we lind from (6), § 138 

tan (Zi /a) = tan (0. - O,) = tan 62 - tan 0^ 
^ ' '^ ^ - /^ 1 + tan O2 tan ^1 

But tan ^1 = wii and tan 0^ — ^2- Hence we have 

As special cases of this relation we obtain the familiar condi- 
tion for parallelism and perpendicularity (§§ 64, ^^^. For, if 
the lines are parallel, (Zi l^ — 0° or 180° ; hence m^ — m^. 
If the lines are perpendicular, (Zi, Zo) = 90° or 270° ; hence 

l-f-mim2 = 0, or mi= • 

mo 

201. Standard Forms of the Equation of a Straight Line. 

We recall here for reference the standard forms of the equation 
of a straight line derived in Chapter III : 

The general equation : Ax -\- By -^ C = 0. 

The slope form : y = mx + b. 

The point-slope form : y — Ui = ^{x — Xi). 

The last two forms are not general, since they will not serve to 
represent lines parallel to the 2/-axis. The first is general. If 
the first represents a line not parallel to the ?/-axis (B =^ 0), it is 
readily reduced to the slope form, by solving the equation for y : 

A C 

y = X — — • 

^ B B 

This yields, as was shown in § 63, 

A 



308 MATHEMATICAL ANALYSIS [XI, § 201 

EXERCISES 

1. Construct a line through the point (— 2, 3) having an inclination 
of 00°. What is the slope ? Write the equation of the line. Find the 
points at which the line crosses the a:-axis and the y-axis. 

2. Proceed as in Ex. 1 for a line passing through the point (2, — 3) 
with an inclination of 135°. 

3. Find, to the nearest minute, the inclination of each of the follow- 
ing lines. Use a table of natural functions. 

(a) 2x-Sy = 0. (c) x = 2.1y + 3.5. (e) x - ?/ + 249 = 0. 
{b) y = OAx-{-l.'I. (d) 7x-\- Sy-8 = 0. (f)x + 2y+6 = 0. 

4. Fhid the tangent of the angle from the first line to the second line 
of each of the following pairs. Then find the angle. 

{a)2x-Sy = 0, (c)x + 3y-3=0, 

ic + 2y+7=0. 3x — ?/ + 6=0. 

(b) 6x + 2y -10 = 0, (d) y = 2x + 3, 
2x-|-32/ + 6 = 0. Sx + y-6 = 0. 

5. Find the equation of the line through (4, 5) and parallel to the 
line joining (—1,2) and (2, — 3). 

6. Find the equation of a line through the intersection of2x + y — 5=0 
and x — 3y+5 = 0, and perpendicular to the line 2x — 3?/-fG = 0. 

7. An isosceles triangle has for its base the line x—2 y+2=0 and for 
its vertex the point ( — 3, 5). The base angles are 45°. Find the equations 
of the other two sides and the coordinates of the other two vertices. 

8. Given the lines aiX + biy + ci = and a^x + 62?/ -{- C2 = 0. Show 
that they are parallel, if and only if 05162 — «26i = ; and that they are 
perpendicular, if and only if aia2 + &162 = 0. 

9. The sides of a triangle have slopes equal to ^, 1, and 2. Show 
that the triangle is isosceles. 

10. Find the angles of the triangle whose vertices are (3, 4), (— 3, 6), 
and (2, - 1). 

11. Find the slope of the bisector of the angle which a line of slope — 2 
makes with a line of slope 3. 

12. The slope of a line AB is 2. Find the equation of a line through 
the origin which makes with AB an angle whose tangent is — 1. 

13. P is any point on the curve whose equation is 2/2 = 4 x. Show that 
the tangent to the curve at P bisects the angle which the line joining P to 
the point (1, 0) makes with the line through P and parallel to the x-axis. 



XI, § 202] THE STRAIGHT LINE 309 

202. The Expression Axi-{-Byi-\- C. The expression x — 2y 
-f 3 has the value + 2 when x = l and y = 1', the value when 
x = —l and 2/ = 1 ; the value — 2 when a; = — 3 and y=l. 
The only interpretation we are able thus far to give to these 
facts is that the second set of values for x and y are the coordi- 
nates of a point (—1, 1) which is on the line whose equation is 
x-{-2y -{- 3 =0, while the other sets of values are the coordi- 
nates of points not on this line. 

It seems reasonable to expect, however, that the value of 
the expression xi—2yi-\-3, where (x^, y^) is any point in the 
plane, must have some relation to the line whose equation is 
x — 2y-\-3 =0. This relation is indeed very simple. * The 
reader should have no difficulty in proving that the value -f 2 
obtained ^bove from the point (1, 1) represents in sign and 
magnitude the directed segment drawn parallel to the a;-axis 
from the line to the point (1, 1). Similarly, the value — 2 
represents the segment drawn parallel to the a>axis from the 
line to the point (—3, 1). 

We proceed to show that a similar result applies to the 
values of the left-hand member of the equation of any line in 
the form Ax^By-\-C=0. 

Let the line I (Fig. 180) be the line whose equation is 
Ax-\-By-{- (7=0, where we assume A=^0, 
and suppose the equation has been 
written so that A is positive. Why is 
this last always possible? The line is 
then not parallel to the aj-axis. Why ? 
Let Pi{xi, yi) be any point in the plane 
and let Q(/i, y^ be the point in which 
the line through P parallel to the a;-axis meets I. Since Q is 
on ?, we have ^i , td , n a 

or By^-\-C = - Ah. 



e;^!^ Pi (x,.v,) 



Fig. 180 



310 MATHEMATICAL ANALYSIS [XI, § 202 

The value of Axi -{- Byi + C, wliicli we are seeking, is therefore 
equal to Axi — Ah, or A{xi — h). But Xi — h represents, in sign 
and magnitude, the segment QPi- We have then, 

Ax^ + By,+ C = A'QP;. 

We conclude that, HA is positive, the number Axi -f Byi + C is 
positive if {xi, 2/1) is to the right of the line Ax-\-By-{-C=0, and 
negative if (xi, 2/1) is to the left of this line. Moreover, Axi + Byi + C 
is proportional to the horizontal distance from the line to the point 

(a^i, 2/i)- 

Finally, if ^ = and B =^ 0, we may suppose the equa- 
tion By-{-C=0 so written that B is positive. The line I is 
then parallel to the a;-axis. Writing its equation in the form 
y =— C/B, it is readily seen that the expression 



-m 



2/. -(t5^ 1 = 2/1+ J 



represents the directed segment drawn parallel to the y-asia 
from the line to the point Pi (Fig. 181). We may then con- 
clude that, B being positive, the number Byi-^-O 
^li^vVi) is positive if the point (x^, y^) is above the line 

1 By + C = 0, and negative if the point {x^, y^ 

is below this line. Moreover, Byi + C is jyro- 

portional to the distance of the point from the 

line. 

Fig. 181 ^ , ,. , ,. . . , 

By the preceding results, we may distinguish 

between the positive and negative sides of a line. If the equa- 
tion of a line is written in the form Ax -\- By -{- C = and so 
that its first term is positive, the right-hand side of the line is 
positive and the left-hand side is negative, unless the line is 
parallel to the aj-axis. In the latter case the upper side is 
positive and the lower side is negative. 



XI, § 203] 



THE STRAIGHT LINE 



311 



T 


-/; 




qA-^t,,., 


/ 


/ 


/^ 


X 




Fig. 182 



203. The Distance of a Point from a Line. The results of 
tlie last article enable us to find the perpendicular distance 
of a point Pi(aJi, y^ from the line whose 
equation is Ax + By -f (7 = 0. If 
A 4=^, the required distance d = MPi 
(Eig. 182) is evidently equal to QPi sin^, 
where 6 is the inclination of the line. 
This is true whether the inclination is 
acute or obtuse, and whether P^ is on 
the positive or negative side of the 
given line. Since 0° ^ ^ < 180°, sin 6 is necessarily positive, and 
d = QP^ sin 6 will have the same sign as QPi ; i.e. it will be 
positive when Pi is on the positive side of the line, and nega- 
tive when Pi is on the negative side. 

We have, from the preceding article, 

QP _ Axi + By, + C 

and, since tan B = — A/B, we have 

sin^ = ^ 



VA^-{-& 
Hence, the required distance is 
(8) MPi=d = '^^'i±^Mi±£, 



If .d = 0, the required distance, by § 202, is simply 

, , C Byi + C 

^^ B B 

But this is precisely what (8) becomes for ^ = 0. Hence (8) is 
true in every case. 

The distance d is positive if (xi, y^ is on the positive side of 
the line, and negative if (a^i, 2/1) is on the negative side, provided 



312 



MATHEMATICAL ANALYSIS [XI, § 203 



the equation is written in the standard form with the first term 
positive. 

Example 1. To find the distance from the line 2a;— S?/— 10 =0 to 
the point (— 3, 1). Since the equation is in standard form the desired 
result is obtained by substituting the coordinates of the given point in the 
left-hand member of the equation and dividing by the square root of the 
sum of the squares of the coefficients of x and y. Hence the distance d 

^® ^^ 2(-3)-5. 1-10 ^-21 

V2'-^-|-(-5)2 V29 

The negative sign indicates that the point is at the left of the line. 

Example 2. Find the equation of the bisector of the acute 
+ 12 = and 4 ic - 3 ?/ + 6 = 0. 

First draw the lines (Fig. 183). 
know from geometry that the bisector of 
an angle is the locus of the points equidis- 
tant from the sides of the angle. Let (x, y) 
be any point on the desired bisector. In- 
spection of the figure shows that (aj, y) is 
on the positive side of one of the lines and 
on the negative side of the other. Hence, 
any point on the desired bisector must 
satisfy the condition that its distance from 
one of the lines is equal to minus its dis- 
tance from the other. This condition is 
expressed by the equation : 



between the lines 3 x 


-4y- 
p 




T7 




A 










/. 


/ 








/y 








y 


p' 








J 








^^ 


/ 








^^'/ 










<-^J- 








7 


- ,' 






X 


-^i 


-f 








- -i^i 


7 




































■■ 























angle 



We 



Fio. 183 



(9) 
or 

(10) 



3 X — 4 j/ -f 12 
5 



4 a- - 3 y -h 6 
5 



18=0. 



7x-7y 

Moreover, any point which satisfies relation (9) is a point of the bisector. 
Hence, we conclude that the equation 7x — 72/ + 18 = Ois the required 
equation. 

NoTK. Had the equation of the bisector of the obtuse angle been 
desired the figure shows that in this case a point on the bisector is either 
on the positive side of both lines or on the negative side of both lines. 
Hence, any such point must satisfy the relation obtained by placing its 
distance from one line equal to its distance from the other line. The 
equation of this bisector is x -f- y 4- 6 = 0. 



XI, § 203] THE STRAIGHT LINE 313 

Example 3. Prove that the locus of a point which moves so that the 
algebraic sum of its distances from any number of fixed lines is constant, 
is a straight line. 

Each of the given straight lines has an equation of the form 
ax+by -\- c = 0. The distance of any point (x, y) from such a line is 

ax -\- by ■^- c ^ 
Va^ + &2 

The equation of the required locus is, therefore, of the form 
aiX -hbiy -\- ci ^ ^ a nX + M + Cn _ q^ 

Since this is an equation of the first degree, the locus is a straight line. 

EXERCISES 

1. Without using a figure determine whether the following points are 
at the right or the left of the line 2x + 3?/ - 5 = 0: (1, 2), (1, - 1), 
(- 2, 1),(1, 1), (4, - 2), (7, - 2), (4, - 1). Then, draw a figure con- 
taining the line and the points and verify the results obtained. 

2. Find the distance of the point (3, —2) from the line 4x— 3y+6=0. 

3. Find the distance of each of the following points from the line 
associated with it. In each case interpret the sign of the result. 

(a) (2,5),4x + 3y-2=0. (e) (- 4, 1), 3^/ - 2 = 0. 

(6) (-3, 7),5a; + 12?/+24=0. (/) (a, a), a: + ?/ - a = 0. 

(c) (2, - 2), 3 a: - 4 y = 0. {g) (6, a),ax + by = Q. 

{d) (5, 2),2x + 5 = 0. (Ji) (1,3), 2/ = 2 a; + 5. 

4. Determine the region of the plane defined by each of the following 
sets of relations, 

(a) a; + 2 ?/ + 4 > 0, (&) 2 x - ?/ + 2 > 0, (c) 2 ic - 3 «/ + 6 > 0, 
a;_2i/-6>0. ?/-2<0. 3a; + 2y-12<0, 

x-y-l<0. 

5. Define by inequalities (as in Ex. 4) the inside of the triangle 
■whose sides are given by the expressions in Ex. 4, (c) equated to zero. 

6. Define by means of inequalities the inside of the triangle whose 
vertices are (- 2, 5), (4, 1), (- 1, 1). 

7. Find the distance between the two parallel lines 3x — 62/ + 5 = 
and 3x- 6?/- 2 = 0. 

8. Find the equation of the bisector of the acute angle between the 
lines 2a; + 32/-4 = 0, x-2y + 7 = 0. 



314 MATHEMATICAL ANALYSIS [XI, § 203 

9. Find the equation of the bisector of the obtuse angle between the 
lines in Ex. 8. 

10. Prove that the bisectors of the angles formed by the two lines 
a\X + 6iy + Ci = and a2X 4- h^y + c^ = are perpendicular to each other. 

11. Find the lengths of the altitudes of the triangle whose vertices are 
(1,2), (-2,3), and (-3, -4). 

12. Find the area of the triangle in Ex. 11 by multiplying half the 
length of one of the sides by the corresponding altitude, and check the 
result by finding the area by the formula of § 196. 

13. Find the distance of the point (1, 2) from the line 3x + 4 ?/ + 12 
= by finding the coordinates of the foot of the perpendicular dropped 
from the point on the line and then using the formula for the distance 
between two points. Check by means of § 203. 

14. If the equations of two parallel lines are ax -\-hy + c z=Q and 
ax 4- by -f c' := 0, prove that the distance between them is the absolute 



value of (c — c')/y/a^ + h'^. 

15. Prove that the bisectors of the angles of a triangle meet in a point. 
[Hint : Choose a convenient relation between the triangle and the 

axes. ] 

16. Find the altitudes of the triangle formed by the lines 

x+2?/-3 = 0, x-y = 0, 4a;-y-l = 0. 

17. Prove that the altitudes on the legs of an isosceles triangle are 
equal. 

18. Prove that the three altitudes of an equilateral triangle are equal. 

19. Prove that the sum of the absolute distances of any point within 
an equilateral triangle from the sides of the triangle is constant. 

204. Two Equations representing the same Line. If of 

two equations of the first degree one can be obtained from the 
other by multiplying the latter by a constant, the equations 
obviously represent the same line, since all the points which 
satisfy one equation must then satisfy the other also. We 
now proceed to prove the converse of this statement : 

If the equations Ax + By + C = and A'x + B'y -{- C = 
represent the same line, either one can he obtained from the other 
by multiplication by a constant. 



XI, § 205] THE STRAIGHT LINE 315 

Let us suppose first that none of the numbers A, A', B, B\ 
C, C is zero. The intercepts of the two lines on the ic-axis are 
then - C/A and - C'/A', on the y-Sixis - C/B and - C'/B'. 
Since the lines are by hypothesis identical, we have 
A A' ^. B B' 

From these relations follow at once 

A _B^_C__. 

A'~B~C'~ ' 

where A; is a constant. It follows that 

A^kA', B = kB', C=kC\ 
If C (or C") is zero, the corresponding line passes through the 
origin, and hence the other line must also pass through the 
origin ; hence C" (or C) is also zero. We leave the rest of 
the proof as an exercise, with the suggestion that the slopes 
of the two lines be compared. 

205. The Intercept Form. Hesse's Normal Form. We 

have called attention thus far to three forms of the equation 
of a straight line : (1) the general equation ; (2) the slope 
form ; (3) the point-slope form. Two other forms are some- 
times of great convenience. These are the so-called mtercept 
form and normal form. The intercept form is 

(11) 1+1='^' ^"^^^'^ 

where a and h represent, respectively, the x- and ^/-intercepts 
of the line. This equation may be derived by finding the 
equation of the line through the points (a, 0) and (0, h). The 
derivation is left as an exercise. (See Ex. 21, p. 89.) This 
form is not applicable if the straight line passes through the 
origin, or if it is parallel to either axis. Why ? 




316 MATHEMATICAL ANALYSIS [XI, § 205 

The normal form is associated with the name of Hesse,* who 
used it extensively. It uses the length p of the perpendicular 
droj^ped from the origin upon the line and the angle a which 
this perpendicular makes with the a:-axis to determine the line. 
To derive the equation when p and a are given, we try to 
find a relation which is satisfied by the coordiDates (x, y) of 
every point P on the line and which is 
not satisfied by the coordinates of any 
other point. To this end (Fig. 184) we 
note that the projection of the broken 
line OMP on the perpendicular OQ is 
equal to p, if and only if P is on the 
line. The projections of the parts OM 
and MP on OQ are, respectively, x cos a and y sin a. The 
desired equation is, therefore, 

(12) X cos a + y sin a = ^ 

We shall take the positive direction of 0$, or p, from the origin 
towards the line, and choose the positive angle XOQ to be a. It is then 
evident that the position of any line is determined by a pair of values of 
p and a, it being understood that p and a are positive and that a is less 
than 360°. 

Moreover every line determines a single positive value oi p and a single 
positive angle a less than 360", unless p = 0. . When p = {) the line evi- 
dently passes through the origin and the above rule for the positive 
direction of p becomes meaningless. When p — 0, it is customary to 
choose a < 180°. 

To reduce the general equation Ax -{- By -\- C =0 to the 
normal form, we need merely observe that in the latter form 
an essential condition is that the coefficients of x and y are 
numbers the sum of whose squares is 1, since sin' a -|- cos^ « = 1. 
We must then multiply all the coefficients of Ax -{-By -^ C = 
by a number k, so chosen that (kAy -^(kBy = 1. This condi- 

♦LuDWiG Otto Hesse (1811-1874), a noted German mathematician. 



XI, §205] THE STRAIGHT LINE 317 

tion will be satisfied if ^ 



Therefore the desired reduction is obtained by dividing the 
equation through by ± V^^ + B^, and transposing the constant 
term to the right-hand side of the equation : 

A B ^ - (7 _ 

The sign of the radical must be chosen opposite to the sign of 
C, or if (7 = 0, the same as that of B. Why ? 

One advantage of the normal form is that every Hne may have its equa- 
tion written in the normal form. Whether the line passes through the 
origin or is parallel to an axis is immaterial. 

EXERCISES 

1. Reduce the following equations to the normal form. Find in each 
case the values of a and p. 

(a) 4a; +8?/- 10 = 0. (d) 3x-2 y + 6 = 0. 

(6) a;-!/ + 5=0. (e)y=2x-3. 

(c) a; + VS ^ = 0. (/) a; = 2y - 5. 

(gr) The equation of the line whose intercepts are — 5 and 2, respectively. 

2. Reduce to the intercept form each of the lines in Ex. 1 for which 
such reduction is possible. 

3. What are the normal formsof the equations x =3,2 a:-f 3 = 0,^—1=0? 

4. Derive the process of reducing the equation Ax + By -}- C = to the 
normal form by using the fact (derived from § 203) that p =— Cj^J A^-\-&. 

5. What system of lines is obtained from the normal form, if a has a 
fixed value, while p is allowed to assume different values ? If p has a 
fixed value and a is allowed to assume different values ? 

6. Find the equations of the lines which pass through the point (1, 2) 
and are two units distant from the origin. 

7. Find the equations of the lines parallel to 5 x + 12 y = 13 and 3 units 
distance from it. 

8. Find the equations of the lines parallel to 3 x + 4 ?/ = 13 and 7 
units distance from it. 



318 MATHEMATICAL ANALYSIS [XI, § 205 

MISCELLANEOUS EXERCISES 

1. Find the equation of the straight line passing through the point 
(3, 4), such that the segment of the line between the axes is bisected at 
that point. 

2. Show that the lines y = ax -{■ a, for all values of a, pass through 
a fixed point, 

3. Given aix + biy + ci = 0, aox + b2y + C2 = 0, agX + bsy + ca = 0, 
the equations of three lines forming a triangle. Show that the equation 
of any line Ax -^ By -^ C = in the plane may be written in the form 

ki{aix + biy + Ci) + koia-zx + b2y + C2) + ks^asx + bsy + ^3)= 0, 
where ki, k2, ks are constants. 

4. Find the ratio in which the line Sy = 6 — x divides the segment 
joining the pohits (6, 1) and (— ^>, 2). 

5. Find the equation of the line that passes through the point (1, 7) 
and makes an angle of 45° with the line x + 2 ?/ = 1. 

6. Find the equation of the line that passes through the point (1, 7) 
and makes an angle of — 45° with the line x + 2y = 1. 

7. Prove analytically that the perpendicular bisectors of the sides of 
a triangle meet in a point. 

8. Prove analytically that the altitudes of a triangle meet in a point. 

9. Prove analytically that the bisectors of the interior angles of a 
triangle meet in a point. 

10. Prove analytically that the bisectors of two exterior angles of a 
triangle and of the third interior angle meet in a point. 

11. Theequationsof two sides of a parallelogram are x— 2 i/=l, x+y=S. 
Find the equations of the other two sides if one vertex is at (0, — 1). 

12. Find the equation of the line passing through the point (1, 1) and 
dividing the segment from (— 7, — 2) to (7, — 1) in the ratio 2:6. 

13. Two vertices of an equilateral triangle are (1, 1) and (4, 1). 
Find the coordinates of the third vertex. There are two solutions. 

14. jdind the equation of the line passing through the point (1, 2) and 
intersecting the line x -\- y = 4^ at a, distance ^VlO from this point. 

15. Find the equation of the line through the point (1, 2) which forma 
the base of an isosceles triangle with the sides 2x — y = 1, x -{- y = I. 

16. A straight line moves so that the sum of the reciprocals of its 
intercepts on the two axes is constant. Show that the line passes through 
a fixed point. 



XI, § 205] THE STRAIGHT LINE 319 

17. If a straight line be such that the sum of the perpendiculars upon 
it from any number of fixed points is zero, show that it will pass through 
a fixed point. 

18. Find the equations of the sides of the square of which two opposite 
vertices are (3, — 4) and (1, 1). 

19. Derive the formula for the distance of a point (ici, ^i) from the 
line-^x i- By + C = hy finding the intersection of the perpendicular 
through the given point and the given line, and then using the formula for 
the distance between two points. 

20. Prove that if the sum of the first moments of n points with respect 
to each of two given perpendicular lines is zero, the sum of the moments 
of tliese points with respect to any line in the plane through the inter- 
section of the given lines is zero. (See Ex. 19, p. 305.) 

[Hint : Take the given perpendicular lines to be the axes of 
coordinates. ] 

21. If with the center of gravity of n points in a plane is associated 
the sum of the weights of the n points, prove that the sum of the first 
moments of the n points with respect to any line in the plane is equal to 
the first moment of the center of gravity with respect to the same line. 

22. Given two half-lines r, s issuing from a point P, a third half-line t 
through P is completely determined if the ratio sin (r^)/sin (ts) = k is 
known. The ratio k is called the simple ratio of t with respect to r, s. 
Prove that the equations I = and m = of r and s, respectively, may be 
so written that, for all positions of t, the equation ot t is I — km = 0. 

23. Given two points Pi(xi, y{) and P'z(x2, yi) and a straight line 
ax -\-hy + c = () which meets the line PiPi in Q. Find the simple ratio 

[Hint : This can be obtained very readily from a figure by observing 
the relation between the desired ratio and the ratio of the distances of 
Pi, Pi from the given line.] 

24. From the last exercise derive the theorem of Menelaus : If a 
straight line cuts the sides of a triangle ABC in three points A\ B', C, 
the product of simple ratios 

AC BA> CB> 
C'B ' A'C' B'A 

is — 1. The point A' is on the side opposite A. B' on the side opposite JB, 
O on the side opposite C. 



CHAPTER XII 
THE CIRCLE 

206. Review. The circle is the locus of a point which moves 
so that its distance from a fixed point, called the center, is con- 
stant. This constant distance is called the radius of the circle. 

If the center of a circle is at the point {h, k) and the radius 
is r, the equation of the circle is 

(1) {X - hy + (1/ - ft)2= r\ 

Tor, this equation expresses directly the fact that the square of 
the distance from the given point (7i, k) to the variable point 
{x, y) is r^. Hence, every point on the circle satisfies this 
equation and, conversely, any point not on the circle does not 
satisfy it. 

In particular, if the center is at the origin (h = 0, k = 0), the 
equation becomes 

(2) x' + y''^ 7-2^ 

"We note also that equation (1) when expanded has the form 

(3) x''-hy^ + Dx + Ey + C = 0. 

It follows that every circle in the plane may be represented 
by an equation of this form. To what extent is the converse 
true? Under what conditions does an equation of the form 
(3) represent a circle ? The answer to this question may be 
obtained by reference to the method of § 183. 

We desire to complete the square on the terms in x, and also on 
the terms in y. Therefore we rewrite the equation in the form 

ix^ + Dx-{- )-\.(y'^-^Ey+ ) = -C. 
320 



XII, § 207] THE CIRCLE 321 

To complete the squares in the two parentheses we need to add 
i)Y4 to the first and J5jy4 to the second; to maintain the 
validity of the equation we must add the same terms to the 
right-hand member. We then obtain 

or 

(:H-f)V(..f)^=- + f--^. 

Since the sum of the squares of two real numbers is positive or 
zero, the left-hand member is positive or zero if x, y, D, E are real 
numbers. Hence the equation can be satisfied by real coordi- 
nates X, y only if L^ -\- E"^ — ^ C i^ di, positive number or zero. 

If Z>2+ ^ — 4 C is positive, equation (3) represents a circle 
with center at (— D/2, — E/2) and radius equal to 



I -y/D^ + E^-4.a 

If i)2_|_ ^2 _ 4 (7 ig zero, equation (3) is satisfied by the coor- 
dinates of the point (— -0/2, — E/2) and by the coordinates of 
no other (real) point. 

If Z)2 4- ^2 _ 4 (7 is negative, equation (3) represents no real 
locus. The answer to our question may then be formulated as 
follows : If (3) represents a curve at all, it represents a circle. 

207. The Equation of a Circle satisfying given Conditions. 

The problem of finding the equation of a circle satisfying 
given conditions resolves itself simply into the problem of 
determining from the given conditions the values of U, Jc, r in 
equation (1), or of D, E, C in equation (3) of § 206. The fol- 
lowing examples will illustrate the methods that may be used : 

Example 1. Find the equation of the circle passing through the three 
points (3, — 5), (3, 1), and (4, 0). 

The desired equation must be of the form (8), and must be satisfied by 

Y 



322 



MATHEMATICAL ANALYSIS [XII, § 207 



the coordinates of each of the three given points. If the first point satis- 
fies this equation, D, E, and C must be such that 

32 +(- 5)-^ + ■i> • 3 + E(-6)+ C=0, 
i.e. such that 

SD-6E+ C=-S4. 

We find similarly from the second and third of the given points, 
3i> + ^+O = -10, 
4j[> + C=-16. 
Solving these three linear equations for Z>, U, C, we obtain 

Z)=-2, E = 4, C = -S. 
The desired equation is, therefore, 

x2 -J- ^2 _ 2 ic + 4 2/ - 8 = 0. 

Another method of solving this problem would be to regard (A, A) as 
unknown coordinates of the center. They must satisfy the two equations 
(3 _ A)2 + (_ 5 _ i.)2 =(3 _ hy +(1 _ j^y2^ 

(4-/i)2+(0- A:)2=(3-/02+(l-^)2. (Why?) 

By solving these equations we can determine h and k. Having found the 
center, it is easy to determine the radius. Then the desired equation can 
be written down in form (1). The completion of the work here sug- 
gested is left as an exercise. What other method could be used to solve 
this problem ? 

Example 2. Find the equation of the 
circle inscribed in the triangle ichose sides 
are y-S=0, 3x— 4?/— 9=0, and 12x+5y 
+ 9 = 0. 

Let (h, k) be the center of the circle. It 
must be equidistant from the three sides. 
The distances of (A, k) from the three given 
lines are — (A: — 3), - ^ (3 A — 4 A: — 9), and 
ji^(12 ft + 5 ^• + 9), the signs being so chosen 
that each of these numbers is positive when 
(ft, k) is within the triangle. (See Fig. 185.) 
By placing the first of these distances equal to tlie second and third, re- 
spectively, we obtain two equations involving ft and k. The solution of 
these two equations yields ft = 1, k — \. Hence the center is the point 
(1, 1). The radius is evidently equal to 2. Why? Therefore the 
required equation is 

(a; - 1)2 +(2/ - 1)2 = 4, or a;2 + y2 _ 2x - 2t/ - 2 = 0. 



" X " 


--Y- 


^ 


zzzzzY~.y-.-.iz 


HrnTNMTffl 


:i=::;z^ii=:=i: 


:: zt ::v= : :: 





Fig. 185 



XII, § 207] THE CIRCLE 323 

EXERCISES 

1. Write the equations of the circles described below: 
(a) Center at the origin, radius equal to 5. 

(6) Center at (1, 2), radius = 4. 

(c) Center at (- 3, — 2), radius = 3. 

(d) Center at (a, a) and radius = a. 

(e) Center at (—2, 1) and passing through the point (3, — 2). 
(/) Center at (2, 1) and tangent to the x-axis. 

2. Discuss fully the locus of each of the following equations : 

(a) x-^ + y^ - 2 X + 4:y -{- I = 0. {d) x'^+ y"^ -\- \ =0. 

(b) x^ + y^ -ix- 6y = 0. (e) x^ + y^ -\-2x- 6y + 10 = 0. 

(c) x^ -\-y^ + Sx-i = 0. (/) x2 4- 2/2 + 2 aa; + 2 a2 = 0. 

(g) Sx^ + Sy^-\-2x-iy-8=0. 

3. What can be said of the coefficients Z>, E, and C in the general 
equation if the equation represents a circle which 

(a) passes through the origin ? 

(6) has its center on the x-axis ? on the y-axis ? 

(c) has its center on the line x -\-y = 0? 

(d) touches both axes ? 

(e) has its radius equal to 2 ? 

4. Find the equations of the circles described below : 
(a) Passing through the points (0, 2), (1, 4), (1, 0). 

(6) Circumscribing the triangle whose sides are the lines «+ y — 3=0, 
x-2y + Q = 0, x + 2 = 0. 

(c) Inscribed in the triangle whose vertices are (0, 2), (0, — 4), and 
(-4,1). 

(d) Having ( — 2, 4) and (4, — 2) as the extremities of a diameter. 

(e) Passing through the points (1, 2) and (2, ]) and having its center 
on the line 2x + y + 2 = 0. 

(/) Tangent to both coordinate axes and passing through the point 
(2, 1). How many solutions are there ? 

5. Prove analytically that any angle inscribed in a semicircle is a 
right angle. 

6. Prove that the locus of a point which moves so that the sum of 
the squares of its distances from any number of fixed points is constant 
is a circle. Find the coordinates of the center of this circle in terms 
of the coordinates of the fixed points. If the number of fixed points is 
three, how is the center of the circle related to the triangle whose ver- 
tices are at the fixed points ? 



324 MATHEMATICAL ANALYSIS [XII, § 207. 

7. Find the equation of the locus of a point which moves so that the 
ratio of its distances from two fixed points is constant and equal to k. 
Determine fully this locus. Examine especially the case k=\. 

[Hint : Let the two fixed points be (a, 0) and (— a, 0)]. 

8. Draw the loci of Ex. 7 for different values of k. Prove that if 
any one of these loci crosses the line joining the two given points in P and 
Q^ respectively, and the raid-point of the segment joining the given points 
is M, we have MP • MQ equal to the square of half the segment. 

208. Tangent to a Circle. Point Form. In § 184 we saw 

how the slope of the curve Ax^ -f- Bi/ + Dx -\- JEy -{- C = at 
any point {xi, y-^ on the curve could be derived. Applying 
this method to the circle 

x'^j^if^Dx-\-Ey+ C = 0, 

we find the slope m at {x^, y-^ on the curve to be 

2xy^ + D 

2y,-\-E 

The equation of the tangent at the point {xi , y^ is, therefore, 

Simplifying, we obtain 

(4) 2x]X-\-2 y^y ^Dx-\-Ey -2 x^^ — 2y{'— Dx^ — Eyi = 0. 

But {xif z/i) is on the curve, and hence 

Xi'-hyi' + Dx, + Ey,-\-C=0. 

If this identity be multiplied by 2 and added to (4) we obtain 

2x,x + 2y,y J^Dx-^Ey + Dx, -j- Ey^ + C = 0, 
or 

(5) x,x + y,y + 1 D{x + x,)-^l E{y + y{)-\-C=.0, 

As a special case of this equation (for D = 0, E = 0, C = 
— r^) we obtain the equation of the tangent to the circle 
x"^ -\- y^= 9-2 at the point (xi, y^) to be 

(6) x,x + y,y = r\ 



XII, § 209] THE CIRCLE 325 

209. Tangent to a Circle. Slope Form. Another form of 
the equation of a tangent to the circle x^ -\- y"^ = r^ is often very 
serviceable. It is derived as follows. The straight line 
y = mx -\- h meets the circle x"^ -\- y"^ = r^ in points whose 
abscissas are given by the equation 

a;2 -I- {mx + hf = r». 

When expanded this equation becomes 

(1 + m2)a;2 -^2mhx -\- h"^ - r'^ = 0. 

The roots of this equation will be real and distinct, real and 
coincident, or imaginary, according as 

is positive, zero, or negative. 

Translated into geometric terms, this means that the line 
y = mx + h will meet the circle in two distinct points, two 
coincident points, or not at all, according as the expression 
above is positive, zero, or negative. If the line meets the 
circle in two coincident points, the line is a tangent. The 
condition ^ ^^^^ - 4(1 + m?) {¥ - r^) = 

yields, after simplification, 

or, b = ± rVl + m\ 

Hence, for all values of m the equation 



(7) y = mx ± r VI + m^ 

represents a tangent to the circle ^ -\-y'^ — r^. 

It follows at once that for all values of m the equation 



y-k= m(x -h)± r VI + m^ 
represents a tangent to the circle (x — hY-\-{y — ky = r^. 



326 MATHEMATICAL ANALYSIS [XII, § 209 

EXERCISES 

1. Write the equations of the tangents to the following circles at the 
points indicated : 

(a) a;2 + ?/2 = 25, at (3, -4). 

(&) a;2 + y2 = 6, at(-l, 2). 

(c) a;2 + ?/2 = 4, at (0, 2). 

(d) jc2 + y2 _ 13^ at the points where x = 3. 

(e) a;2 4- ?/2 = 10, at the points where y = 1. 
(/) a;2 + 2/2 + 2 a: - 4 y = 0, at (1, 1). 

2. Derive the equation of the tangent to the circle (x — h)^ + (y — k)^ 
= r2 at the point (xi, yi) by making use of the fact that the tangent is 
perpendicular to the radius through the point of contact. 

3. Find the intersections of the following circles with the lines in- 
dicated : 

(a) x2-|-y2_ 5andy = 3a;-|-5. (c) ic2+?/2=:i3and3 a:+2 y-13=0. 

(6) x2 + r/2 = 25 andx - 2 !/ - 5 =0. (c?) x^ + 2/2 = 10 and y = 3 x + 10. 
(e) x2 -f ?/2 = 4, and y=-2x + 4, ?/ = -2x + 2 a/5, y = — 2 x + 5. 

Draw a careful figure showing the circle and the three lines. 

4. Write the equations of the tangents to the following circles, the 
slopes of the tangents being as indicated. Find the points of contact. 

(a) x2 + 1/2 = 10, slope = - 3. {d) x2 + 2/2 ^ 25, slope = 0. 

(6)x2 + 2/2 = 5, slope = ^. (c) (x-l)2+(2/+2)2=10,slope=3. 

(c) x2 + 2/2 = 13, slope = |. 



5. Will the equation y = mx ± rVl + m2 represent any tangent to the 
circle x2 + 2/2 = r2. Why ? 



6. What is the point of contact of the tangent 2/ = wix + rVl + w2 
to the circle x2 + y- — r^'i From this result derive the equation 
a^ia: + y\y = r2. 

7. Any circle through the origin has an equation of the form 
x^+y^+Dx-{-Ey=0. Why? Prove that the equation of the tangent at 
the origin is Dx-^Ey=0. This may be done in at least two different ways. 

8. Prove analytically that from an external point two real tangents 
can be drawn to a circle. 

9. Derive the equation y = mx±r\/l -\- m^ directly from the property 
that a tangent to a circle is perpendicular to the radius through the point 
of contact. 



XII, §211] THE CIRCLE 327 

210. The Value and Sign of the Expression x^ + y^ + Dx^ 

+ Eyi + C. The left-hand member of the standard equation 
(x — hy +{y — ky = r^ represents the square of the distance 
from the point (x, y) to the point (h, k). Hence the expression 

(8) (^x^-hy+{y,-ky-r^ 

is positive, negative, or zero according as (xi, y{) is outside, in- 
side, or on the circle whose equation is (x — hy -\-{y — ky = r^. 

Moreover, from Fig. 186 it follows that if (xi, 2/1) is a point 
outside the circle, the expression (8) 

is equal to the square of the length ^ ^{x^,Vj) 

of a tangent drawn from the point 
(^'d 2/1) to t^® circle. Since the left- 
hand member of the general equation 
x^ -\- y"^ -\- Dx -{- Ey -\- C = may be writ- 
ten in the form (x — hy -\- {y — ky — r^ 

we may conclude that the sign of the 

-^ ^ Fig. 186 

expression x^ -\- y^ + Dx^ -f- Ey^ + (7 is 

positive or negative according as the point (x^, y^ is outside or 

inside the circle x'^-{- y"^ -\- Dx -\- Ey -f C= ; and, if positive, it 

represents the square of the length of a tangent drawn from the 

point (xi, y{) to the circle. 

211. The Equations of the Tangents from an External 
Point. Suppose we desire to find the equations of the tan- 
gents drawn from an external point {xi, yi) to the circle 
a;2 -f- 2/2 _ ^2^ Three methods will be discussed: 

Example, Find the equations of the tangents drawn from the point 
(4, - 3) to the circle x^ + y^ = 5. 

Method 1. Let {xi, y{) be the point of contact of one of the tangents. 
The equation of the tangent at this point is XiX -\- yiy = 6- However, 
since this tangent passes through the point (4, — 3) we have 

(9) 4 xi - 3 yi = 6. 




328 MATHEMATICAL ANALYSIS [XII, § 211 

But the point (cci, yi) is on the circle x^ -}- y2 _ 5^ Therefore ' 

(10) xi^ + yi'' = 5. 

Solving equations (9) and (10), we find the points of contact to be (2, 1) 
and ( — 2/5, — 11/5). Therefore the required tangents are2ic + y— 5 = 
and2a; + lly + 25 = 0. 

Method 2. From § 209 it follows that any tangent (not parallel to 
the ?/-axis) to the circle x^ -\- y^ = 6 is of the form y = mx ± VsVl + m^. 
Since this tangent is to pass through the point (4, — 3) we have 

— 3 = 4 m ± VoV 1 -j- W'^5 
which simplifies to 11 m- + 24 m + 4 = ; th is giv es m =— 2, or — 2/11. 
Substituting these values in y — mx ± VSVl + m^ and simplifying we 
have 2x + ?/-5 = and 2 x + 11 y + 25 = 0. 

Method 3. The equation of any line through the point (4, — 3) is of 
the form ?/ + 3 = m(x — 4). Eliminating y between this equation and 
x2 + y2 — 5 ^e have 

(11) (wi2 + l)x2+ x(-8m2-6m) + (16m2 + 24m + 4)=0. 

Now since we desire y + 3 = w(:k — 4) to be tangent, equation (11) must 
have equal roots, i.e. (- 8 m2 - 6 m)2 _ 4(m2 + 1) (16 m2 + 24 m + 4) = 
or 11 m2 + 24 m + 4 = which gives m=— 2, or —2/11. Therefore 
the equations of the tangents are 2x + ?/ — 5 = and 2x+lly+25 = 0. 

212. The Polar of a Point with respect to a Circle. Let 

us apply the first method of § 211 for finding the equations of 
the tangents from an external point to a circle, to the general 
problem of finding the equations of the tangent from the point 
(xi, 2/1) to the circle x^ -\- if- = r^. The coordinates {x\ y') of the 
point of contact are then found by solving simultaneously the 
pair of equations x'x^ + y'yi = r^, x'^ -f- y'^ = r\ The first equa- 
tion expresses the fact that the point (xi, y^ is on the tangent 
x^x -{- y'y = r"^ ', the second, that {x\ 2/') is on the circle. 

This shows that the straight line XiX + yiy = r^, where (xi, y{) 
is any external point, meets the circle in the points of contact 
of the tangents drawn from {xi, y-^. In other words, 

(12) iCiic + ;viy = 7-2 

is the equation of the line joining the points of contact of the 



XII, § 212] 



THE CIRCLE 



329 



tangents through (xi, 2/1), if the latter point is outside the circle. 
If this point is on the circle, we know that (12) is the equation 
of the tangent at the given point. Finally, if (xi, y-^ is inside 
the circle, (12) represents a definite straight line determined 
by the point and the circle. This straight line (12), whether 
(^1) 2/1) is outside, on, or inside the circle, is called the polar of 




Fig. 187 



(s^ij 2/1) with respect to the circle. The polar of (a^i, y^ with 
respect to a circle is then a uniquely determined line for every 
point (aji, 2/1) in the plane, except the center of the circle. 
(See Fig. 187.) Why this exception ? 



EXERCISES 

1. Are the following points inside, outside, or on the circle x^ + y^ 
-2x + 6y-15=.0? (1,2), (1,0), (1,4), (-3,0), (3,0), (0,2), 
(5, 1). For the points outside, find the length of the tangents drawn to 
the circle. Draw carefully a figure to illustrate each of your results. 

2. What is the length of the tangents drawn from (1, 1) to the circle 
whose equation m 2 r^ -\- '2, y'^ -{- ^ x — b y — \ = Q? 

[Caution : The equation is not in the standard form.] 

3. Find the equations of the tangents drawn from the following points 
to the circle indicated : 

(a) (- 2, 4) ; a;2 + y2 = 10. (d) (3, 2) • x'^ + y^ = 4. 

(6) (5, _ 1) ; x2 + y2 = 13. (e) (4, 3) ; x2 + y2 = 16. 

(c) (3, - 1) ; x2 + 2,2= 2. (/) (7, 1) ; x2 + 2,2 ,, 25. 

4. Find the equations of the tangents drawn from (0, 4) to the circle 
aj2 + y2 _ 2a; + 6y-15 = 0. 

6. Show that the polar of a point P with respect to a circle is per- 
pendicular to the radius or radius extended through the point P. 



330 MATHEMATICAL ANALYSIS [XII, § 212 

6. Show that if P is inside the circle, the polar of P is wholly outside 
the circle. 

7. Show that if the polar of P with respect to a circle whose center is 
cuts the line OP in Q, then OP • OQ=r^, where r is the radius of the circle. 

[Hint : Let the center O be the origin and the line OP the x-axis.] 

8. Show that if the polar of a point with respect to a given circle is 
given, the point is uniquely determined. 

[Hint : This follows directly from the results of Exs. 5 and 7 ; or it 
may be proved directly by identifying the given polar ax-\-by + c = 
with the equation Xix + y\y = r^. In the latter case we should have 
Xi/a = pi/b = — r^/c, which determines xi, yi uniquely.] 

9. A straight line is drawn through a given point P, cutting a given 
circle in the points A and B. Calculate the length of the segments PA 
and PB. Let P be chosen as origin and the line through P and the 
center of the circle as ic-axis. The equation of the circle is then x^ + y'^ 
4- Dx 4- C = 0. If p is one of the segments PA or PB and « is the 
angle which PA makes with the x-axis, the coordinates of J. or P are 
{p cos a, p sin a). Since this point is on the circle we have the equation 

(/) cos a)2 + {p sin ay + !>/) cos « + O = 
for determining the two values of p. This equation reduces to 

p2 -f Z) cos a . /) + = 0. 
It may be noted that the product of the roots pip^ of this equation is C, 
i.e. independent of a. What theorem of elementary geometry does this" 
prove ? Prove also that the product PA • PB is positive or negative 
according as P is outside or inside the circle. 

213. The Intersection of Two Circles. Given two circles 
0^2 -f 2/2 4- D.a: + ^i2/ 4- Ci = 0, 
and ic2 4- 2/2 + D.p: + E^ + 0^= 0. 

The coordinates of the points of intersection are found by 
solving the equations simultaneously. Subtracting the equa- 
tions, we have 

(A-A)a5+(^i-^2)2/ + c,-a = o. 

Every point common to the two circles will satisfy this last 
equation, which is the equation of a straight line. Therefore 
the problem of finding the points of intersection of two circles 




XII, § 214] THE CIRCLE 331 

is equivalent algebraically to that of finding the intersections of 
a straight line and a circle. This problem leads essentially to 
the solution of a quadratic equation in one unknown. There- 
fore we may conclude that two circles may intersect in two 
distinct points (two real roots), may be tangent to each other 
(coincident roots), or may not intersect at all (imaginary roots).* 

214. Orthogonal Circles. Two circles which intersect at 
right angles are said to be orthogonal. In this case the tan- 
gents to the two circles at a point of 
intersection must be perpendicular, and 
the two tangents pass respectively through 
the centers of the circles (Fig. 188). The 
condition for orthogonality is then simply 

that the sum of the squares of the radii _ .„^ 

Fig. 188 

of the circles shall be equal to the square 

of the distance between their centers. If the centers are 
(7i(7ii, ^i) and €2(712, k^ and the radii are Vi and r^ respectively, 
the condition for orthogonality is 

If the equations of the circles are 

,. ox ^' + y' + ^^^ + ^^y + c'l = o» 

^ ^ x^ + y'^ + D,x + Eiy+C2 = 0, 

this condition becomes (see § 206) 

4 4 4 "^ 4 ' 

which when simplified gives 

A A + A^2 - 2(Ci + C2) = 0. 

* The reasoning above breaks down, if Z)i - 2)2 = and Ei — E2 = 0, that is 
when the circles are concentric. In this case, unless C\ —0^ = also (in which 
case the two circles coincide), the two equations are inconsistent and have no 
common solution, real or imaginary. 



332 MATHEMATICAL ANALYSIS [XII, § 215 

215. Pencil of Circles. Let the left-hand members of the 
equations (13), § 214, be represented by Mi and Mz respectively. 
Let us consider the locus of the equation 
(14) Mi-kM2 = 0, 

where k is an arbitrary constant. This equation may, iik^l, 
be written in the form 

(16) x^ + 2,^ + -L3^^» + -L_^^2/+-Y-r = 0. 

which represents a circle for each value of k{=^l). When 
k = l, equation (14) represents the straight line 
(16) (Z>i - D,)x + {El - E,)y + Ci - C^ = 0. 

The system of circles obtained by giving different values to 
A;, is called the pencil of circles determined by the two given 
circles. The straight line (16) is called the radical axis of 
the two given circles, and of the pencil. 

The following properties of a pencil of circles are readily 
proved : 

If the tivo gwen circles intersect in two points A and B, every 
circle of the pencil passes through A and B. 

If the two given circles are tangent to each other at a point Ay 
all the circles of the pencil are tangent at A. 

Tf trough any point in the plane not on the radical axis of the 
circles passes one and only one circle of the pencil. The proofs of 
these theorems are left as exercises. 

Further properties of pencils of circles will be found in the 
following exercises. 

EXERCISES 

1. Find the coordinates of the points of intersection of the following 
pairs of circles : 

(o) x2 + y2 ^ 5 and a;2 + y2 ^ 2 X - 4 2/ + 1 = 0. 

(6) x2 + y2 _ a; _|. 2 J/ = and x^ + y2 4. 2 x — 4 y = 0. 

(c) x2 + y2 4. 2 X - 17 = and x2 + t/2 _ 13 _ q. 



XII, § 215] THE CIRCLE 333 

2. Write the equation of the radical axis of each pair of circles given 
in Ex. 1. 

3. Prove that the tangents drawn from any point of the radical axis 
of two circles to the two circles are equal. 

4. Prove that the circles x^-{-y'^-\-6x — 2y + 2 = and x'^ +y'^ + 4y 
+ 2 = are tangent to each other. Find their point of contact and the 
equation of their common tangent. 

5. Find the equation of the circle through the intersections of the 
circles x^+y^ — 4x — 4 = and x^ + y^+2x — Qy — 2 = and the point 
(3, 3). [It is not necessary to find the intersections.] 

6. Prove that the following circles are orthogonal: x^-\-y^ — 2x—4=0 
and x^+y'^—Q ?/+4=0. In general for the circles : x^ -\- y'^ -\- Dx — C = 
and x^ + y"^ + Ey + C = 0. 

7. Determine C so that x^ + y^ — 2x + 4:y — S = and x^ -\-y^ -\-2x 
+ (7 = are orthogonal. 

8. Prove that the locus of the centers of the circles of a pencil is a 
straight line perpendicular to the radical axis of the pencil. 

9. Prove that if the radical axis of a pencil of circles is chosen as the 
y-axis and the line of centers as the ic-axis, the equation of any circle of 
the pencil is of the form x^ + y'^ + kx -\- C = 0, where C is the same for all 
circles of the pencil ; and that all circles obtained by varying k in this 
equation are circles of the same pencil. 

10. The circles of the pencil in Ex. 9 intersect in distinct points, are 
tangent to each other, or do not intersect at all, according as C is negative, 
zero, or positive. In case (7 = 0, all the circles of the pencil are tangent 
to one another at the origin. Draw carefully three figures, illustrating 
the three kinds kinds of pencils here indicated. 

11. Find the equation of a circle wbich is orthogonal to two given 
circles of the pencil in Ex. 9. 

[Hint : Let the two given circles be 

x"^ + y^ + kix + C=0 and x^ +y^ + kix + C = 0, 

and let the required circle be x"^ + y'^ + D^x + E^y + (^2 = 0. If this circle 
is to be orthogonal to each of the given circles we must have (§ 214) 

B^kx - 2((7 + (72) = and D^iki - 2(0 + (^2) = 0. 
These equations give 2)2 = and d^-C. Hence the required equation 
is x2+y2_|_^2?/— C=0. This yields two remarkable results : (1) The coeflB- 
cient E2 is undetermined, and by varying E^ we have a pencil of circles 
each of which satisfies the condition of being orthogonal to the two given 



334 



MATHEMATICAL ANALYSIS [XII, § 215 



circles. (2) The equation found is independent of An, and k2- Hence, 
every circle of the pencil just found is orthogonal to each of the circles of 
the given pencil. Writing I for E2 to obtain uniformity of notation, wre 
have found two pencils of circles : 



and 



x2 + 2/2 4- A:x + O = 
«2 + 2/2 + ;y _ o = 0, 



such that every circle of either pencil is orthogonal to each circle of the 
other pencil. These two pencils of circles are said to form an orthogonal 
system. (See the adjacent figure.)] 




12. In an orthogonal system of circles, the centers of the circles of one 
pencil are on the radical axis of the other pencil. 

13. If the circles of one pencil of an orthogonal system intersect in two 
distinct points A and J5, the circles of the other system do not intersect at 
all, but pass between the points A and B. 

14. If the circles of one pencil of an orthogonal system are mutually 
tangent to each other at a point A, the circles of the other pencil are also 
mutually tangent at A. 

15. Prove that the three radical axes of three circles (not belonging to 
the same pencil) taken two by two intersect in a point. This point is 
called the radical center. Show that it is the center of a circle orthogonal 
to each of the three given circles and that the tangents drawn from it to 
the given circles are equal. 



XII, § 215] THE CIRCLE 335 

MISCELLANEOUS EXERCISES 

1. Find the condition that ax -{- by -\- c = be tangent to the circle 
«2 + y2 _ r2. 

2. Find the equation of the circle passing through the points (0, 0), 
(a, 0), and (0, &). 

3. Show that the equation of the circle having the points (a^i, yi) and 
(^2, 2^2) as the extremities of a diameter is (x— Xi)(x — a^a) + (.V — Vi) 
(y - 2/2) = 0. 

[Hint : The circle is the locus of the vertex of a right angle whose 
sides pass through the given points. ] 

4. Find the equation of a circle w^hich is tangent to the lines a; = 0, 
y = 0, and ax + by -\- c = 0. 

5. A line is drav^^n through each of the points (a, 0) and (—a, 0), 
the two lines forming a constant angle d. Find the equation of the 
locus of their point of intersection. 

6. A straight line moves so that the sum of the perpendiculars drawn 
to it from two fixed points is constant. Show that it is always tangent to 
a fixed circle. 

7. Give a geometrical construction for the polar of a point with 
respect to a circle. 

8. If the polar of a point P passes through Q, then the polar of Q 
passes through P. 

9. Find the equations of the common tangents of the circles x^+y^=S 
and x2 + ?/2 _ 10 ic + 20 = 0. 

10. Find the locus of a point which moves so that the length of a tan- 
gent drawn from it to one given circle is k times the length of a tangent 
drawn from it to another given circle. 

11. Find the equation of a circle through the points of intersection of 
3:2 +2/2 _ 4 and x^-\-y^—2x-\-4:y+4:=0 and tangent to the line x—2y=0. 

12. Show that the polars of a given point P with respect to the circles 
of a pencil pass through a fixed point, unless P is on the line of centers. 

13. A point moves so that the sum of the squares of its distances from 
the sides of a given square is constant. Show that its locus is a circle. 

14. A point P moves so that its distance from a fixed point A is always 
equal to k times its distance from another fixed point B. Show that its 
locus is a circle, if k =^1. Show also that for different values of k 
these circles have a common radical axis. 



336 MATHEMATICAL ANALYSIS [XII, § 215 

16. A line rotating about a fixed point meets a fixed line in a point 
P. Find the locus of a point Q on OP such that OP • OQ is constant. 

16. Prove that among the circles of a pencil there are at most two which 
are tangent to a given straight line (unless all the circles are tangent to 
the line) . When is there only one ? None ? 

[Hint : Let the given line be the sc-axis.] 

17. Inversion with Respect to a Circle. Given a circle with center 
and radius r. Corresponding to any point P in the plane (distinct from 0) 
there exists a unique point P' on OP such that OP • OP' = r^. The 
point P' is called the inverse of P with respect to the given circle. Prove 
the following propositions : 

(a) If P' is the inverse of P, P is the inverse of P'. 

(b) If P is inside the given circle, P' is outside ; and vice versa. 

(c) Every point on the given circle corresponds to itself. 

(d) If the coordinates of P and its inverse P' are (x, y) and (x', y') 
respectively, referred to two rectangular axes through 0, we have 

x'=-J^, y' = -'^; and x= ^^^' , y= ^^^^ • 

(e) If a point P describes a curve, the inverse P' describes a curve 
called the inverse of the former curve. The inverse of any straight line 
through is this line itself. 

(/ ) The inverse of any line not through is a circle through O, and 
the inverses of parallel lines are circles tangent at 0. 

(g) The inverse of any circle is a circle, unless the given circle passes 
through 0, in which case its inverse is a straight line. 

(h) Two orthogonal circles or lines have orthogonal inverses, 
(i) Any circle orthogonal to the given circle is its own inverse. 

(j) The adjoining figure illustrates a 
simple mechanism for changing circular 
motion into rectilinear motion. It is known 
as the inversor of Peaucellier. The heavy 
lines represent rigid bars, hinged at their 
extremities. The sides of the quadrilateral 
ABCD are all equal and OB - 0D= p. 
Prove that if O.is fixed and the mechanism 
is allowed to move in any way it can, C is 
always the inverse of A with respect to a circle with center O and radius 
r = \JP--p^, where I is the side of the rhombus ABCD. Hence, if A de- 
scribes a circle through 0, G will describe a straight line. 




CHAPTER XIII 
THE CONIC SECTIONS 

216. Definition of a Conic. A conic section* or simply a 
conic is defined as the locus of a point which moves so that its 
distance from a fixed point, Fi, is always equal to a given 
constant, e, times its distance from a fixed line D^D^^. 

The fixed point F^ is called the focus. The fixed straight 
line A A' is called the directrix. The constant e is called the 




eccentricity. It is assumed that e > and that F^ does not 
lie on A A'. 

If P (Fig. 189) is any point on the curve, we have, by 
the preceding definition, 
(1) F,P = e' MP, 

where MP is the perpendicular distance of P from the 
directrix. It must be remembered that F^P and MP are 
absolute quantities, not directed quantities, and that e is 
positive. 

♦ The name "conic section " is due to the fact that the curves in question 
were originally obtained as the sections of a right circular cone. They 
were discussed from this point of view by the ancient Greeks. 
z 337 



338 MATHEMATICAL ANALYSIS [XUl, § 217 

217. The Equation of a Conic. Let the directrix be chosen 
as the 2/-axis and the line through Fi perpendicular to the 
Y directrix as the aj-axis (Fig. 190). The coordi- 

nates of Fi may then be taken as (|), 0), 

y *' where p is different from zero. Let P (a;, y) 

be any point on the conic. Then 



A' 



P F^P.o)X 



F,P = ^{x-py + y% 
Fig. 190 ^^^ ^^ = + a; or - a; 

according as x is positive or negative. Equation (1), § 216 

then becomes 

V(a;-i))2 + 2/2 = ±ea;. 

Squaring both sides of this equation and simplifying, we have 

(2) (1 - e2)a;2 + 3/^ - 2px-{-p'^ = 0. 

This is the equation of the conic. For, the coordinates of 
every point {x, y) satisfying the definition of the conic will 
satisfy equation (2), and conversely, every point whose coor- 
dinates satisfy equation (2) will satisfy equation (1). Why ? 

This is an equation of the type considered in § 183. It 
represents an ellipse if 1 — e^ > 0, a hyperbola if 1 — e' < 0, 
and a parabola if 1 — e'^ = 0. Hence we have, 

A conic is an ellipse, a parabola, or a hyperbola according as 
the eccentricity e is less than 1, equal to 1, or greater than 1. 

THE ELLIPSE 

218. Standard Equation of the Ellipse : e < 1. We have 
seen in § 183 how to determine the locus of equation (2) 
by completing the square. If we apply the same method 
here, equation (2) may be written in the form 

(3) 






(1-e^-- 



XIII, § 218] 




THE ELLIPSE 




or 
(4) 


X — 


P T , 2/' _ 


pH^ 


ri _ p2\ ' 1 _ p2 


^1 — ^2\2 



339 



Since 1 — e^ is positive by hypothesis this equation represents 
an ellipse whose center is at the point {p/{l — e^), 0), and 
whose axes coincide with the two y\ 
straight lines x = p/(l — e^) and y = 
(Fig. 191). 

Let us move the curve parallel 
to the a;-axis through a distance 
-p/(l - e2), i.e. to the left if p > 0. 
Then its center comes to the origin, 
and its equation becomes 




Fig. 191 



(5) 
or 
(6) 



x^ + 



p^e 



2p2 



1-e^ 

y2 



(1 - e^y 



p^e^ 



p2^2 



= 1. 



If we place 



(1 - e2)2 1 



(7) 



(1-62)2 



= «-, 



j9%2 



= 6S 



the equation of the ellipse in its new position, i.e. with its 
center at the origin (Fig. 192), becomes 




Fig. 192 



(I.) 



a2 &2 



From (7) we have 

(8) 62=^2(1-62), 

which shows that b <a, since e < 1. 
If the ellipse is given in the form 
(IJ, a and b are known. Then the 



340 MATHEMATICAL ANALYSIS [XIII, § 218 

value of e can be found in terms of a and b bysolving equation 

(8) ; this gives 

(9) e^ = ^^A^ 

219. Properties of the Ellipse. It is important to distin- 
guish between the properties of a curve as such and those 
properties which are concerned merely with the relations the 
curve bears to the coordinate axes. Thus the ellipse, as a 
certain kind of curve, is symmetrical with respect to two 
perpendicular lines called the axes of the curve. The longer 
of the segments on these lines cut off by the curve is called 
the majon axis, the shorter one, the minor axis. The inter- 
section of the two axes of the curve is called the center of 
the ellipse. 

Every ellipse, no matter how it is situated in the plane 
of coordinates, has a major axis and a minor axis as well as a 
center. From the way in which the equation was derived, we 
know also that every ellipse has a focus and a directrix. The 
symmetry of the curve with respect to the y-'dxis shows that 
this same curve could have been obtained from a second focus 
F2 and a second directrix D2D2 on the opposite side of the 
center. 

We shall now investigate how the two foci and the two 
directrices are related to the major axis, the minor axis, and 
the center. 

220. Foci and Directrices. The original position of the 
focus Fi was (p, 0) ; the abscissa of its new position is 

^ 1 _ e2 1 _ e2 

Since from (7) we know that pe/{l — e^) = a, we find the 
coordinates of the focus F^ in the new position to be ( — ae, 0). 



XIII, § 221] 



THE ELLIPSE 



341 



(See Fig. 193.) Similarly the equation of the directrix AA' 
in its new position is 



X = 



P 



or 
(10) 



__a 
e 



The second focus F2 has the 
coordinates (ae, 0). The second 
directrix D2D2 has the equation 




FiQ. 193 



(10') 



a 

X=~' 

e 



221. The Ellipse in Other Positions. If the center of 
the ellipse is at the origin and the major axis is on the y-axis, 
the equation of the ellipse is 



(I.) 



62 ^ a2 ' 



where, as before, 2 a is the length of the major axis and 2 6 is 
the length of the minor axis. The foci of this curve are at the 
points (0, ae), (0, — ae) ; the equations of the directrices are 

2/ = ± « A- 

The equation of an ellipse whose center is at the point 
(h, k) and whose axes are parallel to the coodinate axes is 

(II.) (£^+(1^=1, (a>6) 

or 

(II„) .(l^ + (L=^^i, (a>b) 



according as the major axis is parallel to the a>-axis or to the 
y-Sixis. Finally we can reduce an equation of the form 
(III) Ax^-hBy^ + Dx-^Ey+C^^O, A>0,B>0, 

to the form II_, or 11^^, if it has a real locus. (See § 183.) 



342 



MATHEMATICAL ANALYSIS [XIII, § 222 



222. The Case a = b. The Circle. If a = b the equation 
(Ij.) reduces to the equation of a circle. The relation a=b 
implies, however, that e = and this value of e is excluded in 
the definition of a conic. On the other hand it is clear that 
for a given value of a, as the eccentricity approaches zero, the 
ellipse approaches a circle. At the same time, the foci ap- 
proach the center, and the directrices recede indefinitely. 
Why ? Since the circle is a limiting form of an ellipse it is 
classified as an ellipse with equal axes and is counted among 
the conies. 

223. A Geometric Property of an Ellipse. An important 
geometric property of any ellipse follows from the fact that 
the distance from the center to either focus, which we shall 
denote by c, is given by the relation 



or 

(11) 



c=:ae= V a^ 



^S 



This relation shows that c, a, and h are the sides of a right- 
angled triangle in which a is the hypotenuse (Fig. 194). In 
other words, a circle drawn with its center 
at an extremity of the minor axis and with 
its 7'adius equal to a, will cut the major axis 
in the foci, Fi and F2. 

In computing the elements of an 
ellipse from a and b, it is generally con- 
venient first to find c from (11) and then 




to find e from the relation* 
(12) 



e = £ 



* This relation is equivalent to (9), § 218. It may be expressed by saying 
that e is the cosine of the angle CF^B, Fig. 194. 



XIII, § 224] 



THE ELLIPSE 



343 



The extremities of the major axis are called the vertices of 
the ellipse. 

The chord through a focus perpendicular to the major axis 
is called a latus rectum. Its length is 2 b^/a. Why ? 



_ 


^ ^ -■■'^= 




W'' _ ^ s 




_, 5^^ 


I ^ Z .5 


•3,22 * ^ 2 no) 


. _C:fi|Q) . J^Q) JC 


S (^ 


^^=, --=^ 


rZ 


i ^ 




±_ » : 



Fig. 195 



224. Illustrative Examples. Example l. Given the ellipse 

4a;2 + 9y2_36 = 0. 

Find the coordinates of the center, the vertices, the foci, and the equa- 
tions^ of the directrices. 

The given equation may be written in 
the form 

9 4 

from which follows that a = 3, 6 = 2. 
Therefore c = Va^ — b^ = \/5 and e = V5/3. 
The coordinates of the center are (0, 0), 
the vertices (3, 0) and (—3, 0), the foci 
(— V5, 0) and ( VS, 0) and the equations of 
the directrices are x = — 9/ V5 and x = 9/ VS 
(Fig. 195). 

Example 2. Find the coordinates of the center, the vertices, the foci, 
and the equations of the directrices of the ellipse 

25 x* + 9 2/2 - 50 x+S6y - 164 = 0. 

From § 183, we know that the given equa- 
tion may be written in the form 

25(x - 1)2 + 9(?/-f 2)2=225, 
or 

(x-1)^ , (y+2)2 ^.^ 
9 25 

We now conclude that the center is at 

(1, — 2), and that the major axis is parallel 

to the y-axis. Here a = 5, 6=3, c = 4, e = | 

Fig. 196 and a/e = ^^. Sketching the ellipse we find 

from the figure that the vertices are (1, 3) 

— 7), and the foci (1, 2) and (1, —6). The equations of the 

33/4. 





^ , " 


~^-4 


-'&" 


^^ ^s^ 


^ 0.2} ^ 


t X 


^" 




(/r2) 




A ^ i 


\ "^^ 


^^^a: 


-._ : ^± -: 


::::i:i±li^:- 



and (1, 

directrices are y = 17/4, y 



344 MATHEMATICAL ANALYSIS [XIII, § 224 

EXERCISES 

In the following ellipses determine the major axis, the minor axis, the 
coordinates of the center, the coordinates of the vertices and foci, and the 
equations of the directrices. Sketch the curves. 

1. 3 a-2 + 4 ?/2 = 12. 7. 3 a;2 + 3 2/2 = 12. 

2. 4 a:2 + 3 ?/2 = 12. 8. x^ + 2y^= 8. 

3. 4x2 + 2/2=16. 9. 4x2 + 9 2/2-16a;-18?/-23=0. 

4. 36 x2 + 25 2/2 = 144. 10. 9 a;2 + 25 2/2 - 150 y = 0. 

5. 2 x2 + 4 2/2 = 3. ^ 11. 4 a;2 + 2/^ - 8 a; + 4 2/ + 4 = 0. 

6. 5x2 + 2/^^=75. 12. 9x2 + 42/2 + 36x-16 2/+16=0. 

13. Write the equation of the following ellipses : 

(a) Center at origin, major axis = 4 on x-axis, minor axis = 3. 

(&) Center at origin, major axis = 5 on 2/-axis, minor axis = 3. 

(c) Center at origin, major axis = 6, minor axis = 3 (two solutions). 

(d) Center at origin, eccentricity 4/5, foci at (—2, 0) and (2, 0). 

(e) Center at (1, 2), major axis = 6 parallel to x-axis, minor axis = 4. 
(/) Foci at (0, 2) and (0, 8), major axis = 10. 

14. An ellipse has its center at the origin, and its axes coincide with 
the coordinate axes. The ellipse passes through the points ( VT, 0) and 
(2, 1). Find its equation. 

[Hint. Assume the equation of the ellipse in the form (1^.). Find a 
and b from the fact that the ellipse must pass through the given points.] 

15. Find the equation of the ellipse symmetrical with respect to the 
coordinate axes if the major axis is twice the minor axis and the curve 
passes through the point (2, 1). How many solutions ? 

16. Show that the equation of the ellipse whose vertex is at the origin 
and whose major axis is on the x-axis is of the form a'^y- = ?>2(2 ax — :e2). 

17. Verify equation (I^) by deriving the equation of a conic whose 
focus is at (— ae, 0) and whose directrix is the line x =— a/e. 

18. Find the equation of the ellipse whose focus is at (0, 0), whose 
directrix is the line x + 2/ — 1 = and whose eccentricity is 1/2. 

19. Find the equation of the ellipse whose eccentricity is 1/3, whose 
focus is at (3, 1) and whose directrix is the line 3x + 42/ — 1=0. 

20. Find the equation of the conic whose focus is at (2, 1), whose 
eccentricity is 3, and whose directrix is the line 3 x + 2/ = 1. What kind 
of a conic is the curve ? 



XIII, § 225] 



THE ELLIPSE 



345 



225. Focal Radii. The segments F^P and F2P joining any 
point P on an ellipse to the foci JF\, F.^, are called the focal 
radii of the point P. 

If the equation of the ellipse is given in the standard form (/,), 
the focal radii of any poi7it P{xi, 2/1) «^e a — ea'i, a + exi. 



A 


Y 


P (x-„v,\ 


D, 


M, 






M, 


r> 


/K 


4 


\^i 


F.J 


X 



Fig. 197 

For, from the definition of an ellipse (Fig. 197), 
• FiP=e'M,P, F.P=e'PM.^ 

But from the figure, we have also 



M,P='^ + x,, 



PM2 = --Xi. 
e 



Therefore the focal radii are 

FiP = a -f ex„ F2P —a — exi. 

From these relations follows the important property : 

The sum of the focal radii of any point of an ellipse is constant 
and is equal to the major axis 2 a. 

It may be noted that this relation still holds when the 
ellipse is a circle (e = 0), although the method of its derivation 
is not applicable in this case. An ellipse could, therefore, be 
defined as the locus of a point which moves so that the sum of its 
distances from two fixed points (the foci) is constant. 



346 MATHEMATICAL ANALYSIS [XIII, § 226 

226. Geometric Constructions of the Ellipse. The property 
of the ellipse derived in § 225 gives the construction indicated 
in Fig. 198 for the points of the ellipse when the foci and 
the major axis are given. 




1 ^~r-^ 

Fig. 198 

The segment AB is the major axis. Different positions of P 
on this segment give corresponding values AP and PB of the 
focal radii of a point on the ellipse. Circles drawn with these 
radii and centers at the foci intersect in points of the ellipse. 
To each position of P on AB correspond four points of the 
ellipse. 

A very convenient method of drawing an ellipse is indicated 
in Fig. 199. Two pins are stuck in the paper at the foci and 




Fia. 199 

a loop of thread thrown over them. If a pencil point is in- 
serted in the loop and moved so as to keep the thread taut, it 
will describe an ellipse. Why ? 

Another method of constructing an ellipse (much used by 
draftsmen) is based on the fact (§ 179) that if the ordinates of 
the circle x"^ + 2/2 = a^ are shortened in the ratio b : a (b < a) 



XIII, § 226] THE ELLIPSE 347 

there results an ellipse with major axis 2 a and minor axis 2 b. 
The adjoining figure (Fig. 200) exhibits the method. Explain 
and prove the method correct. 




Fig. 200 

[Hint. The two circles being of radii b and a respectively, we have 
OB/OQ = b/a ; hence, MP/MQ = b/a. Why ?J 

EXERCISES 

1. Construct an ellipse whose foci are 2 inches apart and whose major 
axis measures 3 inches. 

2. Construct an ellipse whose major and minor axes are 2 and 1.6 
inches respectively. 

3. From the property of § 225 derive the equation of an ellipse. 

4. From Fig. 200 show that the coordinates (x, y) of any point on the 
ellipse (Ij.), p. 339, are given bj'^ the equations 

x= acosQ, y = b sin 0, 

where 6 is the angle MOQ. Do these values of x, y satisfy the equation 
of the ellipse for all values of ^ ? 

5. From the relation between the ordinates of a circle and an ellipse 
whose major axis is equal to the diameter of the circle prove that any 
plane section of a circular cylinder is an ellipse, provided the plane of 
section is not parallel to an element of the cylinder. 

6. Prove from the result of the last exercise that a properly determined 
plane section of an elliptic cylinder is a circle. 



348 



MATHEMATICAL ANALYSIS [XIII, § 227 



THE HYPERBOLA 

227. Standard Equations of the Hyperbola. If e > 1, then 
1 _ e2 < 0, and it is convenient to write (2), § 217, in the form 



(13) 



{f — V)x^ — 'f-\-2jix-f'^ 0. 



Completing the square and transforming as in § 218, we 
obtain / \2 2 ^^ra. 

This equation represents a hyperbola whose center is at the 
point ( — i)/(e^ — 1), 0) and whose axes coincide with the lines 
x = - j>l{f - 1), and 7/ = (Fig. 201). 




\:. 


a 


Xi 






\ 6 
a >y 


A 


^n 


y. 


/" 


K 


\v^\(fl^^o)*x 


/\ 


•>[ 




X 



Fig. 201 



Fig. 202 



If the curve is moved parallel to the a>-axis so that its center 
coincides with the origin (Fig. 202), its equation becomes 



y 



^2g2 



p^e^ 



= 1. 



(e2_i)2 e2_i 
If, then, we place 

the equation of the hyperbola becomes 



(Ix) 






XI II, § 2271 THE HYPERBOLA 349 

' Erom (14) we have the relation connecting a, b, e as 

62 = a2(e2-l), 
or 

(15) e^ = ?i±*?. 

Here, as in the case of the ellipse, it is important to note 
some of the properties of the curve. It is seen that the 
locus is symmetrical with respect to the line passing through 
the focus and perpendicular to the directrix. This line is called 
the principal axis and the segment of this line intercepted by 
the curve is called the transverse axis and its length is 2 a. 
The extremities of the transverse axis are called the vertices, 
and the point midway between the vertices is called the center. 
The curve is also symmetrical with respect to the line through 
the center and perpendicular to the transverse axis. The seg- 
ment on this line whose length is 2 6 and whose mid-point is 
at the center of the hyperbola is called the conjugate axis. 

If a hyperbola has its center at the origin, and if its trans- 
verse axis 2 a is on the ?/-axis, and its conjugate axis is 2 b, its 
equation is 

(I) ^-y-=-i. 

The equation of a hyperbola whose center is at the point 
(h, k), whose transverse axis is 2 a, and whose conjugate axis 
is 2 b, is 

(IIJ (^_(J^ = 1, or (II.) (l|^-(l^^ = -l, 

according as the transverse axis is parallel to the x-axis or the 
y-axis. 

The equation of any hyperbola with axes parallel to the 
coordinate axes may be written in the form 
(III) Ax^-\-By^'-hDx + Ey-\-C=X), A>0,B<0; 



350 



MATHEMATICAL ANALYSIS [XIII, § 227 



and every equation of this form (A > 0, B < 0) represents a 
hyperbola or a pair of straight lines (cf. § 183). 

As in the case of the ellipse, it is easy to show that every 
hyperbola has two foci on the transverse axis, one on each 

side of the center and at a distance 
c from the center, where 
c2 = 02^2 = a2 -I- bK 

With each focus is associated a 
directrix perpendicular to the trans- 
verse axis and at a distance a/e 
from the center (Fig. 202). 

The latus rectum, i.e. the chord 
through the focus and perpendicular to the transverse axis pro- 
longed, is of length 2 b^/a. The asymptotes of the hyperbola 




Fig. 202 (repeated) 



(x-hy {y-ky ^ 



¥ 



= 1 



are the lines 
(16) 



{^-hY 



&2 



228. Geometric Properties of the Hyperbola. Tlie segment 
from the center to a focus of a hyperbola is the hypotenuse of 
a right-angled triangle ivhose legs are the semi-transverse and 
senii-coiij agate axes. Why? It is readily seen, moreover, 
that, if a rectangle be constructed by drawing lines through 
the extremities of each axis parallel to the other axis, the 
diagonals (extended) of this rectangle are the asymptotes of 
the hyperbola (Fig. 202). The circle drawn on either diagonal 
as a diameter passes through the foci. Why ? 

229. Illustrative Examples. 

Example 1. Find the coordinates of the center, the vertices, and the 
foci, and the equations of the directrices and the asymptotes of the hyperbola 
4x2-9.^2 + 36 = 0. 



XIII, § 229] 



THE HYPERBOLA 



351 



The equation is readily transformed into the form 

9 4 

It is now seen that the center is at the 
origin and that the transverse axis is along 
the y-axis (Fig. 203). The vertices are 
(0,2) and (0, -2). Since c = V^, the 
coordinates of the foci are (0, Vl3) and 
(0, — Vi3). The asymptotes are given 
by 4a;2_9y2_o or 2x -Sy = and 
2x-\-Sy = 0. Since e = \/l3/2 the equa- 
tions of the directrices are 



y = ± 



Vis 



±±vrs. 









ss^ F. (h, ^) >j/ 






i j^X'Z^t M 


s / T 


^o^^ 1 ? 


D^ ^i--=^r ?i 


^i^^ ^2^^N 


<^^ ^--'^ ^^^^ 


y^ t-Ta-^tS^ 









Fig. 203 



Example 2. Find the coordinates of the center, the foci, and the ver- 
tices, and the equations of the asymptotes and the directrices of the 

hyperbola 

16 x2 - 9 1/2 4- 32 X + 54 ?/ - 209 = 0. 

The given equation may be written in 

the form 

16(x + l)2-9(y-3)2=144, 

or 

(a; +1)2 (i/-3)2 _.^ 

9 10 

The center is therefore at the point 
(—1, 3) and the transverse axis is parallel 
to the X-axis (Fig. 204). Since a = 3, the 
vertices are (2, 3) and (-4, 3). More- 
over, since c = V9 + 16 = 5, the foci are 
at the points (4, 3) and (- 6, 3). Like- 

14 4 

wise, e = c/a = 5/3 and hence the directrices are x = — — , x = - • The 

asymptotes are given by 

16(a;+ 1)2 -9(2/ -3)2 = 0. 

Why? That is, the asymptotes are the lines 
4x-Sy-\-lS = 0, 

and 

4x-f-3y-5 = 0. 



Y 


\ ite "^ 


^v 4^ 


_ Cv- :---= ^^ 


^^^' ^^L 


^t" M 


A ' /ji\ 


/ \ '^ / \ 


._S7_.X 




\ ~/^\ uJii- 


3j::_ _vt/ 


^Z ^ X2 X 


7/^Z__::^g. " 


^V ^ i^^^S 


Z^ M ^^ \^ 


7 ^ 









Fig. 204 



352 MATHEMATICAL ANALYSIS [XIII, § 229 

EXERCISES 

For each of the following hyperbolas determine the transverse axis, the 
conjugate axis, the coordinates of the center, the coordinates of the ver- 
tices and the foci, and the equations of the directrices and asymptotes. 
Sketch the curves. 

1. Sx^-4y^=12. 7. - 9 x2 + 2/'^ = 36. 

2. 4 x2 - 3 2/2 = 12. 8. 2/2 - 2x2 = 4. 

3. 4x2-3?/2z=_ 12. 9. 4x2- 12?/2_8x- 242/ - 56 = 0. 

4. 3 x2 - 4 2/2 = - 12. 10. 5 x2 - 4 ?/2 -f 10 X + 25 = 0. 

5. _36x2+25?/2 = 144. 11. 9x2 - 16i/2 + 18x - 96y - 279 = 0. 

6. x2 - t/2 = 1. 12. x2 - 2/2 4. 2 X - 2 2/ = 2. 

13. Write the equations of the following hyperbolas : 

(a) Center at origin, transverse axis = 6 on x-axis, conjugate axis = 4. 
(&) Center at origin, transverse axis = 8 on y-axis, conjugate axis = 10. 

(c) Center at origin, transverse axis and conjugate axis = 4, axes 
coinciding with coordinate axes. Two solutions. 

(d) Center at origin, focus at (5, 0) and transverse axis = 8. 

(e) Center at origin, transverse axis = 8, focus at (0, 5). 
(/) Center at Origin, focus at (5, 0), conjugate axis = 8. 

(g) Center at (1, 2), transverse axis = 6 parallel to x-axis, conjugate 
axis = 4. 

(A) Center at (0, 3), focus at (0, 5), conjugate axis = 2V3. 
(i) Foci at (1, 2) and (1, — 8), transverse axis = 6. 

14. A hyperbola has its center at the origin and its axes on the 
coordinate axes; it passes through the points (0, VS) and (2, 3). Find 
its equation. 

[Hint. Since one point of the hyperbola lies on the ^/-axis, the equation 
may be assumed in the form I^^, i.e. 

62 a2 
and a and b may then be determined.] 

15. Show that the equation of any hyperbola whose vertex is at the 
origin and whose transverse axis is on the x-axis is of the form a'^y- = 
hH2 ax -}- .x2). (See Ex. 16, p. 344.) 

16. A hyperbola whose asymptotes are at right angles is called rectan- 
gular. Prove that the equation of a rectangular hyperbola may be written 
in the form x2 — w2 = cfi. 



XIII, § 231] 



THE HYPERBOLA 



353 



230. Focal Radii of the Hyperbola. If P{xi, y^) is any 
point on the hyperbola whose equation is 

i^2 _ ^ ^ 

the focal radii F^P and F2P are given by the equations 
F^P = exi + a, F^P = ex^ — a. 

The proof of the above statement is left as an exercise. It 
is analogous to the corresponding proof ^n the case of the 
ellipse (§ 225). 

Hence, the difference of the focal radii of any point on a hyper- 
bola is a constant. 

A hyperbola could, therefore, be defined as the locus of 
a point which moves so that the difference of its distances from 
two fixed points (the foci) remains constant. 

231. Conjugate Hyperbolas. Any hyperbola determines 
uniquely a second hyperbola whose transverse and conjugate 
axes coincide in position and length with the conjugate and trans- 
verse axes respectively of the first 
hyperbola (Fig. 205) . Thus, if the 
equation of the first hyperbola is 



a2 



62 



1, 



the equation of the second hyper- 
bola is ^2 2/2 



x^ 
a2 




Fig. 205 



Each of the two hyperbolas thus related is called the conjugate 
of the other, and the two hyperbolas are called conjugate 
hyperbolas. 

Two conjugate hyperbolas have the same asymptotes. Why ? 
2a 



354 MATHEMATICAL ANALYSIS [XIII, § 231 

EXERCISES 

1. Geometric construction of the hyperbola. Show how to construct 
a hyperbola given the foci and the length of the transverse axis by a 
method depending on the property of the hyperbola derived in §230 
and entirely analogous to the first method described in § 226 for con- 
structing the ellipse. 

2. Derive the equation of the hyperbola from the definition suggested 
at the end of § 230. [Let the foci be i^i(c, 0) and i^2(- c, 0) and let 
the constant difference of F\P and F2P be 2 a.] 

3. What is the equation of the hyperbola x"^ — y^ — a^ after it has 
been rotated about the origin through an angle of 45° ? (Cf. § 190.) 

4. From the result of Ex. 3 determine the length of the transverse axis 
of the hyperbola xy = k. 

5. What are the equations of the hyperbolas conjugate to the hyper- 
bolas in Exs. l-12,p. 352? 

6. Prove that the foci of two conjugate hyperbolas are on a circle. 

THE PARABOLA 

232. Standard Equations of the Parabola. If in § 217 we 

let e = 1, equation (2) becomes 

(17) y^-2px+p^ = 0. 
or 

(18) y' = 'ip(^-fi- 

We saw in § 183 that this equation repre- 
sents a parabola whose vertex is at the 
point (p/2, 0) and whose axis coincides 
with the line y = (Fig. 206). If the curve is moved parallel 
to the ic-axis so that its vertex coincides with the origin, the 
equation of the curve becomes 

The focus of the curve is now at the point (i>/2, 0) and its 
directrix is the line x = — p/2 (Fig. 207). 




Xlir, § 233] THE PARABOLA 355 

The following theorems follow directly. Their proofs are 
left as exercises. 

The equation of a parabola whose vertex is at the origin 
and whose axis coincides with the ?/-axis is 

(I,) x^=2/.j,. 

The equation of a parabola whose vertex 
is at the point {li, k) and whose axis is 
parallel to the a;-axis is 

(II,) (y-ky = 2p(x-h). 



Fj(±l\o) X 



Fia. 207 



The equation of the parabola whose vertex is at the point 
(h, k) and whose axis is parallel to the 2/-axis, is 

(II,) {x-hy = 2piy-k). 

The equation of any parabola whose axis is parallel to the 
cc-axis is of the form 

(III.) By'^ + Dx-\-Ey^C = 0. 

The equation of any parabola whose axis is parallel to the 
2/-axis is of the form 

(in J Ax^JrDx-\-Ey-\.C=0, 

The distance from the vertex to the focus and from the 
directrix to the vertex of the parabola y^ ==2px is p/2. 

233. Geometric Properties of the Parabola. The chord 
drawn through the focus and perpendicular to the axis is 
called the latus rectum. Its length is twice the distance from 
the focus to the directrix. 

The focal radius connecting any point P{xi, 2/1) on the parabola 
y2= 2px to the focus is equal to Xi -{-p/2. 

The proofs of these properties are left as exercises. 



356 



MATHEMATICAL ANALYSIS [XIII, § 234 



234. Illustrative Examples. Example l. Given the parabola 
r^ = 6 y. Find the coordinates of tlie vertex and 
the focus, and the equation of the directrix. 
Sketch the curve. 

The vertex is at (0, 0) and the axis of the 
curve coincides with the y-axis (Fig. 208). The 
distance from vertex to focus is 3/2. Therefore 
the focus is at (0, 3/2). Likewise, the distance 
from vertex to directrix is 3/2. Hence the equa- 
tion of the directrix is y =— 3/2. To sketch the 
curve, mark the focus, draw the latus rectum and 
then sketch the curve. 



1 






_:_ _ + ^_ 


\ : 


V ~-t 


L A 








-^ -"""*:- 


y ^ 











Fig. 208 



Example 2. Given the parabola ?/2=:— 8x+2?/+15. Find the coordi- 
nates of the vertex and the focus, and the 
equation of the directrix. Sketch the curve. 

The given equation may be written as 

(y-l)2=-8(a:-2). 

Therefore the vertex is at (2, 1) (Fig. 209), 

and the axis is parallel to the a!;-axis. The 

distance from vertex to focus and from 

directrix to vertex is — 2. Therefore tlie 

focus is at (0, 1) and the equation of the 

directrix is x — 4t. The curve is readily 

sketched by plotting the focus and marking 

off the latus rectum. It may also be sketched by plotting another point 

or two. 



















1 








'H 




~ 








'*' 


s 








I 


























s 
































s 
































\ 
























































































(0, 


























































1 




J 








\: 






















/ 




























/ 




























/ 




























/ 




























^ 














T 


/ 










X 


' 





















































Fig. 209 



EXERCISES 

Sketch each of the following parabolas. Determine the coordinates 
of the vertex and the focus, and the equation of the directrix. 



1. y2 = 4a., 

2. y^ =—4x. 

3. 2/2 = 4 ic + 2. 



4. y2__4a._^2. 

5. x2 = 4 y. 

6. x2=-4y. 



10. a;2 -}- 4 X — 4 y + 6 = 0. 

11. 2/2 _ 2a; -4?/- 8 = 0. 

12. a:2 4- r/ -(- 1 = 0. 



7. x^ = 4y +2. 

8. x2 =-4?/ + 2. 

9. 2/2 = 6x+ 12. 



13. 2/2 = - 4 X + 2 y -f- 8. 

14. y'^ + 2x-4y = 0. 
16. x2 - 2 X -}- 2 y = 0. 



XIII, § 235] THE PARABOLA 357 

16. Write the equation of each of the following parabolas : 
(a) Vertex at (0, 0) and focus at (2, 0). 

(6) Vertex at (0, 0), axis coinciding with ?/-axis, curve passing through 
the point (8, 4). 

(c) Focus at (— 1, 3) and directrix the line a; — 1 = 0. 

(d) Vertex at (1, — 2), axis parallel to x-axis, distance from vertex to 
focus equal to 2. 

(e) Vertex at (0, 2) , directrix parallel to a:-axis and parabola passing 
through the point (2, 1). 

235. The Intersections of Conies and Straight Lines. The 

coordinates of the pomts of intersection of the ellipse 

(19) 6V -f ay = a'^b^ 
and the straight line 

(20) y = ma; + k, 

are found by solving these two equations simultaneously for 
(x, y). Eliminating y, we obtain the quadratic equation 

(21) {W + a}m')x^ + 2 a'mkx + a\k''- - h^) = 0, 

the roots of which are the abscissas of the points of intersection. 
For each of these roots the corresponding ordinate is found by 
substituting in (20). Why not in (19) ? We accordingly ob- 
tain, in general, two solutions {x, y). These solutions are real 
and distinct, real and equal, or imaginary, according as 

(22) 62'^ (j2^2 _ ]^2 ^0, =0, or < 0. 

Corresponding to these three cases, the straight line intersects 
the ellipse in two distinct points, in two coincident points {i.e. 
in a single point), or not at all. 

The discussion just given includes for a = 6 the case of the 
intersection of a circle and a straight line. 

To treat the intersection of the hyperbola ly^x^ — a V = «^^^ 
with the straight line (20), we need only notice that alge- 
braically we can reduce this problem to the preceding by 
simply writing — 6^ for h"^. Why ? 



358 MATHEMATICAL ANALYSIS [XIII, § 235 

This leads to the equation 

((^2^2 _ 52)3.2 ^ 2 a^mkx + a2(fc2 + b"^) = 0. 
This is a quadratic equation unless a^mS _ 52 _ q^ if ^^2^2 _ 52 
= 0, the line (20) is parallel to an asymptote, and, ii k ^ 0, it 
meets the hyperbola in only one point. If A: = the line is an 
asymptote and does not meet the curve at all. If a'^m'^ — b^z^O, 
we conclude that the line (20) intersects the hyperbola in two 
distinct points, two coincident points {i.e. in only one point), or 
not at all, according as 

(23) k'^-o?m^-\-¥>0, =0, < 0. 

Finally, the line (20) will meet the parabola 

(24) y' = 2px, 

in the points whose abscissas are the roots of the equation 

m2a;2 + 2(mk - p)x -\- k^ = 0. 

If m = 0, the line meets the curve in only one point. If m ^ 0, 
the line will intersect the parabola in two distinct points, two 
coincident points, or not at all, according as 

(25) p-2mk>0, =0, OY < 0. 

Similar results are evidently secured also for straight lines 
x = k, parallel to the 2^-axis. We then have the theorem : 

Any conic is met by a straight line in the plane of the conic in 
two distinct points, a single point, or not at all. 

EXERCISES 

1. Draw figures illustrating all the results of the last article. 

2. In a manner similar to that of the last article discuss the intersec- 
tions of the line y = mx -^ k and the conic y'^ = 2px-~ gx^. 

3. Derive conditions analogous to (22), (23), and (25) of the last article 
when the straight line is assumed in the form Ax + By -\- C = 0. These 
conditions are slightly more general than those given in the text. Why ? 



XIII, § 236] THE PARABOLA 359 

236. Tangents and Normals. Slope Forms. If, for a given 
value of m, the valae of k in the equation y = mx -f k is so deter- 
mined that the intersections of the line y = mx -\- k with a given 
conic coincide, i.e. so that the quadratic equation determining 
the abscissas of the points of intersection has equal roots, the 
line will be tangent to the conic. Why ? (See § 209.) 
The slope forms of the equations of the tangents to a conic 
result directly from the middle one of each of the conditions 
(22), (23), and (25) for the determination of k. Hence the 
equation of the tangent whose slope is m is : 
for the ellipse b'^x'^ -f a^y^ = a^b^^ 



(26) y =mx± Va^m^ + b^ 

for the hyperbola b'^x'^ — a^y^ = a^b"^, 



(27) y = mx± Va^m^ - b^ ; 
for the parabola y^ == 2 px, 

(28) y^mx+J--. 

Am 

We note that for a given slope the parabola has only one 
tangent, the ellipse two, and the hyperbola either two or none 
according as ahn^ — b^ '^ or < 0. [The condition a^m^ — &^ 
= yields the asymptotes.] 

The line drawn perpendicular to a tangent through its point 
of contact P is called the normal at P. 

EXERCISES 

1. Find the equations of the tangents to the following conies satisfying 
the conditions given, and find for each tangent its point of contact : 
(a) 4x2+ 9y2 = 36, m = ^. 
(6) y^ = Sx, inclination 30°, 45°, 135°. 
(c) 9 x2 - 25 2/2 = 225, perpendicular to x + y -{-I =0. 
{d) x^ -y^ = 1, parallel to 5 x -f 3 y — 10 = 0. 
(e) y^ = 8x, perpendicular to 2x — Sy + Q = 0. 



360 MATHEMATICAL ANALYSIS [XIII, § 236 



2. Show that the line y = mx ± Vb'^ — a'^m^ is tangent to the hyper- 
bola &2a;2 — a2y'2 -f a2^2 _ Q for all real values of m lor which h^ — d^nf- > 0. 

3. For what value of k will the line y =2x + k be tangent to the 
hyperbola x2 -4 2/2-4 = 0? 

4. Find the coordinates of the points of intersection of the line 
3x — w + l=0 and the ellipse x^ + 4 y2 = 55. 



6. Find the points of contact of the tangents y = mx± Va^m^ + b'^ to 
the ellipse b'^x'^ + a^y^ = a^b'^. 

6. From the result of Ex. 5 find the equations of the normals to the 
ellipse &2x- + a'^y'^ = a'^b'^ wliose slope is m. 

7. By the method suggested in Exs. 5 and 6, find the equation of the 
normal to the hyperbola bH^ — a-y^ = a^b'^ in terms of its slope. 

8. Same problem as Ex. 7 for the parabola y^ = 2 px. 

9. A tangent to the ellipse b'^x:^ -\- a^y^ = a-b^ will pass through the 
point (xi, yi), if yi = mx\ ± Va'^m'^ + b'^. By solving this equation for m 
show that through a given point (xi, yi) will pass two distinct tangents, one 
tangent, or no tangents, according as b'^Xi^-^a^yi^—a^b'^ > 0, =0, or < 0. 

10. By the method of Ex. 9, discuss the number of tangents that can 
be drawn from a given point (xi, yi) to the hyperbola b'^x^ — a^y^ = cfib'^', 
to the parabola y'^ =2 px. 

11. Find the equations of the tangents to the parabola y^ = ix which 
pass through the point (—2, — 2). 

12. Find the equations of the tangents to the ellipse 4 x^ + y2 = ig 
which pass through the points ( V3, 2) ; (0, 4) ; (0, 8) . 

237. Tangents. Point Form. The slope of the curve 

(29) Ax^ + By^ + Dx + Ey-\-C= 0, 

at a point (a^i, 2/1) on the curve, was found in § 184 to be 

2By,-\-E 
Hence the equation of the tangent to (29) at {xi, yi) is 

2/ — 2/1 = — — -— ^ — (» — ^i)' 
This reduces to 

(30) 2 Axix + 2 Byiy + Dx-\-Ey = 2 Ax^'^ + 2 By^^ -f Dx^ -f Eyi 



XIII, § 237] THE PARABOLA 361 

Since (xi, yi) is by hypothesis on the curve (29), we have 

2 Ax,^ 4- 2 Byi^ = -2 Dx^-2 Eyi-2 C. 

Substituting this value in the right-hand member of (30), 

2 Axx^ + 2 Byiy -\- Dx -\- Ey = - Dx^ - Ey^ -2 0. 

Hence, by transposing and dividing by 2, we obtain the equation 
of the tangent to (29) at the point (x^ y^ in the form 

(31) Ax,x + By,y + 2)(^^±^ + E^^^^ + C = 0. 

A A 

This equation is readily written down from (29) by replacing 
a;2, 2/2, X, and y by x^x., y^y, ^{x + x^), and }(y + y^), respectively. 
By applying this rule to the standard equations of the conies 
which are special cases of (29) we obtain : 

The equation of the tangent at the point (xi, y^ 



to the ellipse « y2 


is 


Jfi^, l/i2/_1. 
a2 + b2-^' 


to the hyperbola 






aj2 y^ ^ 
a2 62 


is 




to the parabola 






2/2 = 2px 


is 


ViV = P{x + Xi 



EXERCISES 

1. Write the equation of the tangent to each of the following conies 
at the point indicated : 

(a) x2 + 4i/2 = 8, at (2, 1). 
(6) 4a:2_3?/2 = 9, at (3, -3). 

(c) y'^ — Qx = 0, at the point where y = — 3. 

(d) x'^-y'^z=zA^ at (2, 0). 

(c) x2-2?/2^_4, at(-2, 2). 

(/) y2 — 4 X = 0, at the extremities of the latus rectum. 



362 



MATHEMATICAL ANALYSIS [XIII, § 237 



2. Write the equation of the normal to each of the conies in Ex. 1 at 
the point indicated. 

3. Find the equation of the normal to each of the conies hH"^ + ahj'^ = 
a^b^ b-x^ - aV = «^&^» and y^ = 2px at the point (xi, yi). 

4. Prove that the tangents drawn to an ellipse at the extremities of any 
diameter (chord through the center) are parallel. 

6. Show that an ellipse and a hyperbola with common foci intersect at 
right angles. 

6. Show that the tangents at the vertices of a hyperbola meet the 
asymptotes in points at the same distance from the center as are the foci. 

7. Find the angle (in degrees and minutes) at which the two curves 
a;2 + 2 y'-^ = 9 and ?/2 + 4 x = intersect. 

8. Show that the secant of the parabola y^ = 2px joining the points 
(xi, yi) and (X2, yt) on the curve has the equation 2 px - (?/i + yi)y -\- yiy2=0. 
Show that this reduces to the equation of the tangent when the given 
points coincide. 

238. Geometric Properties of Tangents and Normals to 
the Parabola. Let the parabola have the focus F, the vertex 

V, and the directrix d, the latter 
meeting the axis VF in D (Fig. 
210). If the vertex is chosen 
as origin of a system of rectan- 
gular coordinates and the axis 
is chosen as the avaxis, while 
the segment DF is denoted by 
p, the equation of the parabola 
is y^ = 2 px. Now let P{x^, 2/0 
be any point on the parabola. The equation of the tangent at 
this point is y^y =p{x -\-x^. This tangent meets the axis of 
the parabola (the a>-axis) in the point T{—x^y 0). Hence 

TV= VM, 

where M is the foot of the perpendicular dropped from P on 
the axis. From this, and by the definition of the parabola, 



d 

L 


V-^^^ 


/^ 


'V 


\ 


T D 


Fig. 21 






XIII, § 238] THE PARABOLA 363 

follow the relations 

TF=DM=LP=FP, 

where L is the foot of the perpendicular drawn from P to the 
directrix. Hence TFPL is a rhombus. We conclude further 
that ZLPT=ZTPF', 

and, if S is the intersection of the diagonals of the rhombus 
TFPL, that the angle FSP is a right angle. Moreover, the 
line drawn through V, the mid-point of TM, perpendicular to 
TM, passes through S. We have then the following theorems : 

Theorem 1. Tlie tangent to a parabola at any point P bisects 
07ie of the angles formed by the focal radius of P ayid the line 
through P parallel to the axis of the parabola ; the normal at P 
accordingly bisects the other angle. 

Theorem 2. The foot of the perpendicular dj-opped from the 
focus on any tangent to the parabola is on the tangent at the 
vertex. 

EXERCISES 

1. Prove theorems 1 and 2 of § 238 analytically. 

2. Give a geometric construction for the tangent to a given parabola 
at a given point. (The axis of the curve as well as the curve is supposed 
to be given. ) 

[A geometric construction means a construction with ruler and 
compass.] 

3. Given the focus and directrix of a parabola, show how any num- 
ber of points of the parabola can be constructed on the basis of the 
results of the last article. 

4. Given the focus of a parabola and the tangent at the vertex, use 
Theorem 2 of § 238 to draw any number of tangents to the parabola. 
These tangents will give a vivid picture of the shape of the curve ; the 
tangents are said to envelop the curve. The curve itself is not supposed 
to be given. 

5. The outline and axis of a parabola are given ; show how to 
construct the focus and directrix. 



364 MATHEMATICAL ANALYSIS [XIII, § 238 

6. To construct the tangents to a giv6n parabola from a given 
external point. Assume that the focus and directrix and hence the axis 

are given. 

[Analysis: If Q is the given point, it 
follows from Theorem 1 of the last article 
that A QLiPi and A QFPi are congruent. 
Hence, 



Li 






^'- 


^ 


^, 


kD 


KX 




l; 


'XpF 

IV 






QLi = QF. 

We determine Zi (and Z2), therefore, as 
the intersection with the directrix of the 
circle with center Q and radius QF. Com- 
plete the construction. How is the con- 
struction affected when Q assumes various positions in the plane ? When 
is the construction impossible and why ? What happens when Q is on 
the curve ? 

7. In the figure of Ex. 6, prove that the line through Q parallel to the 
axis bisects the "chord of contact" P1P2. 

8. If a parabola is rotated about its axis the sur- 
face generated is called a paraboloid of revolution. 
Prove that if a source of light is placed at the focus 
of such a paraboloid*, all the rays issuing from the 
source will be reflected in the same direction (par- 
allel to the axis of the paraboloid). This is the prin- 
ciple of the so-called parabolic reflectors, used on searchlights, etc. 

9. By an argument similar to that employed in § 212, prove that the 
equation of the chord of contact of the tangents drawn from an external 
point (xi, 2/1) to the parabola y'- = 2px is ijiy =p(x + Xi). This line 
is called the polar of the given point with respect to the parabola. It is 
defined by its equation even when no tangents can be drawn through the 
given point. 

10. Prove that the polar of a point Q is parallel to the tangent at 
the point in which the line through Q parallel to the axis meets the 
parabola. 

11. Prove that the length of the so-called subnormal My oi a parabola 
at the point P (see Fig. 210) is independent of the position of P on the 
curve. 

12. Prove (Fig. 210) that TF = FN = FP and that FS=\ FN. 

♦The focus of the generatmg parabolai is called the focus of the paraboloid. 



XIII, § 238] 



THE PARABOLA 



365 



13. Use the relation FN' = FP (Ex. 12) to show how to construct the 
normal at a given point P of a parabola (the focus and axis also being 
given). Construct a considerable number of normals in this way and 
show that they envelop a curve. (See Ex. 4 for the meaning of 
" envelop.") 

14. Show that any two perpendicular tangents to a parabola intersect 
on the directrix. 

239. Geometric Properties of Tangents and Normals to 
the Ellipse. If for any ellipse we let the coordinate axes coin- 
cide with the axes of the curve, the equation of the ellipse has 
the form 

The equation of the tan- 
gent at any point Pi(a;i, i/i) is 

b^x^x -h a%y = a^b^. _ 

A 

The a>-intercept (Fig. 211) 
of this tangent is 

Xi 

The remarkable thing about 

this result is the fact that it is independent of b and of y^. 
This means that if any other ellipse be given having the axis 
A' A in common with the first ellipse, then the tangent drawn 
to this new ellipse at a point having the abscissa x^ will also 
pass through T. This is therefore true of the circle drawn on 
A' A as diameter. If A' A is the major axis of the ellipse, this 
circle is called the major circle of the ellipse ; similarly the 
circle drawn on the minor axis of any ellipse as diameter is 
called the minor circle. 




* We do not in this article impose the restriction a > 6. 



366 MATHEMATICAL ANALYSIS [XIII, § 239 

A geometric construction for the tangent at any point Pj of 
an ellipse follows readily from the above considerations (as- 
suming that in addition to the curve one of the axes is given). 
Figure 211 shows the construction using the major circle and, 
in broken lines, the construction using the minor circle. 

The following theorem is of fundamental importance in dis- 
cussing the geometric properties of the ellipse : 

Theorem 1. Tlie tangent and the normal to an ellipse at a given 
point bisect the angles formed by the focal radii drawn to the point. 

Proof. We are to prove that the tangent at Pj (Fig. 212) 
bisects the angle F^PxR, and that the normal at P^ bisects the 
angle F^P^F^. To this end we calculate first the tangent of 
the angle SP^R. Using the equation of the ellipse as given 
above and taking the foci to be PgCc, 0) and Pi(— c, 0), we have 

the slope of the tangent P^S = - ^, 

a^y^ 

the slope of F,R (i.e. FiP,) = -^ — 



The tangent of the angle <^i from P^S to P^R is then 

Xi -\- c a^yi 



tan d)i — — 

^' 1 b%y, 



a%{x, + c) 



Simplifying this expression, we find 
tan <^i = — -. 

The tangent of the angle <^2 from P^S to P^F^ may evidently 
be obtained by simply changing c to — c in the last result. 
(Why?) Hence, ., 
tan <f)2== • 



XIII, § 239] 



THE PARABOLA 



367 



We conclude that <^2 = — ^v This proves that PiS bisects the 
angle F^PiB. That the normal bisects the angle FiP^F^ 
follows at once from elementary geometry. 

The theorem just proved leads at once to another geometric 
construction for the tangent (and normal) to an ellipse at 
a given point, supposing the foci of the ellipse are known. 

Theorem 2. The foot of the perpendicular dropped from 
either focus on any tangent to an ellipse lies on the major circle. 

Proof. (See Fig. 212.) Let S be the foot of the per- 
pendicular dropped from F^ on the tangent P^S, and let it 
meet the line F,P^ in R. Then F^^P^B 
is an isosceles triangle (why?) with 
P^R = P.F^. We have then 

F,R = F,P, + P,F^ = 2 a. (§ 225) 

Also aS' is the mid-point of F2R and 
O is the mid-point of FiF^. Hence 
OS=^FiR==a, and S is on the major 
circle. 

We should note also that, if Q is any 
point on the tangent PiS, then QR = QF2, which is important 
in connection with the problem of drawing the tangents to an 
ellipse from an external point. (See Ex. 5, below.) 




Fig. 212 



EXERCISES 

1. Show how to construct the tangent to a given ellipse at a given 
point. (Two constructions, one using the major circle, one using the 
foci.) 

2. Show that, in Fig. 211, OA is a mean proportional between OM 
and OT. 

3. Show that, in Fig. 212, OFi is the mean proportional between the 
intercepts on the cc-axis of the tangent and normal at Pi. 



368 MATHEMATICAL ANALYSIS [XIII, § 239 

4. Prove analytically that S (Fig. 212) is on the major circle. 

5. Show how to construct the tangents to an ellipse from a given ex- 
ternal point Q. [Hint : Construct B (Fig. 212) as the intersection of two 
circles, one with center Fi the other with center Q.] 

6. Show that if a right angle moves with its vertex on a given circle 
and one of its sides passing through a fixed point within the circle, the 
v-^ther side will envelop an ellipse. 

7. Use the result of Ex. 6 to construct a considerable number of tan- 
gents to an ellipse, given the major circle and one focus (the outline of 
the ellipse is not supposed to be given in advance, but will appear vividly 
after this problem is solved). 

8. If an ellipse is rotated about its major axis the surface generated is 
called a prolate spheroid. Show that sound waves issuing from one focus 
will be reflected by the surface to the other focus. This principle is used 
in the so-called ' ' whispering galleries. ' ' 

9. By an argument similar to that used in § 212 show that the equation 
xxi/a^ -f yyi/b'^ = 1 is the equation of the line joining the points of contact 
of tangents drawn from (xi, yi) to the ellipse x^/a''^ + y'^jW- — 1. 

240. Geometric Properties of the Hyperbola. Many of 
the geometric properties of the hyperbola are similar to cor- 
responding properties of the ellipse, which is to be expected 
in view of the similarity of their equations. The following 
two theorems are fundamental : 

Theorem 1. Tlie tangent at any point of a hyperbola bisects 
the angle betiveen the focal radii di-aion to the point. The normal 
bisects the adjacent supplementary angle. 

Theorem 2. The foot of the perpendicular dropped froin 
either focus on any tangent to a hyperbola is on the circle drawn 
on the transverse axis as diameter. 

The proofs of these theorems are left as exercises. They 
are similar to the proofs of the corresponding theorems on the 
ellipse. Draw figures illustrating Theorems 1 and 2. 



XIII, § 240] THE PARABOLA 369 

Certain new properties of the hyperbola relating to the 
asymptotes will be found among the exercises below.. 

EXERCISES 

1. Show how to construct the tangent and the normal to a given 
hyperbola at a given point. 

2. If P is any point on a hyperbola, OA the semi- transverse axis, 
ifcf the foot of the perpendicular dropped from P on OA (produced), and 
T the point in which the tangent at P meets OA, prove that OA is a 
mean proportional between Oilf and OT. 

3. With the notation of Ex. 2 show that OFi is the mean propor- 
tional between ON and OT, Fx being the focus on OA and N the point 
in which the normal at P meets OA (produced). 

4. Prove Theorem 2 (§ 240) analytically. 

5. Show how to construct the tangents to, a hyperbola from an ex- 
ternal point. 

6. Show that if a right angle moves with its vertex on a given circle 
and one of its sides passing through a fixed point outside the circle the 
other side will envelop a hyperbola. 

7. The construction of tangents to a hyperbola analogous to Ex. 7, 
p. 368. 

8. Use Ex. 3 above and Ex. 3, p. 367, to show that an ellipse and 
hyperbola having the same foci intersect at right angles , 

9. Prove that, if a tangent to a hyperbola meets the asymptotes in 
Ti and T2, the point of contact of the tangent is the mid-point of the 
segment T1T2. 

10. Prove that the area of the triangle formed by any tangent and the 
asymptotes of a hyperbola is constant (= a6). 

11. Show that if a straight line cuts a hyperbola in Pi and P2 and the 
asymptotes in ^1 and ^2 the segments PxQx and P2^2 are equal. Use 
this result to construct any number of points of a hyperbola when the 
asymptotes and one point of the curve are given. 

12. By an argument similar to that used in § 212 show that the 
equation xxx/a^ — yyi/h^ = 1 is the equation of the line joining the 
points of contact of the tangents drawn from (cci, 2^1) to the hyperbola 
a;2/a2 - 2/2/62 = 1. 

2b 




370 MATHEMATICAL ANALYSIS [XIII, § 241 

241. The Conies as Plane Sections of a Cone. We stated 
in § 216 that the ellipse, hyperbola, and parabola could all be 
obtained as the plane sections of a right circular cone. This 
we shall now proceed to prove. In doing so we shall get the 
machinery for solving problems of a more general type. 

If a point P in a plane a (Fig. 213) is joined to a point 
S not in a by a straight line SP, the intersection P' of SP 

by a plane a' is called the projection 
of P from S upon a'. Similarly, if 
all the points of a curve in a be 
joined to S, the intersections of these 
lines with a plane a' form a curve C", 
Fig. 213 ^^v/ which is called the projection from S 
of the curve C. The point JS is called the center of projection, 
and the process described is called central projection, to dis- 
tinguish it from orthogonal projection previously considered 
{e.g. in § 135). 

If, now, the curve C in the plane a is a circle, the lines 
through jS and the points of this circle form a cone with vertex 
S. This is not a right cone, in general. As the lines through 
S are not supposed to terminate in S, we get a so-called co7n- 
plete cone, or cone of two nappes, which consists of two con- 
gruent ordinary cones placed vertex to vertex so that their 
axes form a straight line. It will now be clear that a plane 
section of this cone is the same as the projection of the circle 
C from the vertex S upon the plane of section. 

We have then reduced the problem to that of finding the 
central projection of a circle. We will solve it by finding the 
relation between the coordinates of a point P in a and the co- 
ordinates of the corresponding point P' in a'. To this end 
(Fig. 214) let be the foot of the perpendicular dropped from 
JS on the line of intersection of the planes a and a'. Let be 



XIII, § 241] 



THE PARABOLA 



371 



the origin and let the line of intersection OF of the two planes 
be the 2/-axis in the system of coordinates in each of the two 
planes. Let the line OX perpendicular to 01^ in a be the 
a^axis in a, and the line OX' perpendicular to OF in a' be the 
ic-axis in a'. Let the angle between the two planes be ; then 
X'OX= 0. Now let P{x, y) be any point in the plane a, and 




let P\x\ y') be the projection of P from S. We seek the 
relation connecting the coordinates x, y, x', y'. 

Draw ST parallel to OX, and represent the length OS by h. 
Then we have 

T0=^, TS= ^ 



sinO 



tand 



We then have from similar triangles 

x' :{TO-i-x') =x:TS, 
y':y = SM^ : SM= TM' : TO. 



If we substitute the values of TO, TS, and TM' (= TO + x'), 
we obtain , 



x' + 



h 



h ' 



sin tan 



372 MATHEMATICAL ANALYSIS [XIII, § 241 

and , , h 



sind 



1^ 

y _Jl_ 

sin B 

Solving these equations for x and ?/, respectively, we have 

_ li cos 6 • x' _ hy' 

^~ ^mO'X' + h' ^"sin^-x' + Zi* 

If these expressions be substituted for x and y in the equa- 
tion of any curve in the plane a, the resulting equation in x' 
and y' will be the equation of the projection of the curve in a'. 
To solve the problem we proposed at the outset, let the curve 
in the plane a be the circle 

x'^-\-y^=a\ 

The equation of the corresponding curve in a' is then 

7^2 cos2 ^ . x'2 -f hhj"^ = tt2 sin2 e-x'^+2 lia^ sin • x' -\- a%\ 

Collecting like terms, we have * 

(/i2 cos2 e-a" sin2 6) x"- + li'y'^ - 2 ha" sin 6 - x' - aVi^ = 0. 

We see at once that this is the equation of a conic. It is an 
ellipse, a parabola, or a hyperbola according as 

/i2cos2^-a2sin2^>0, = 0, or < 0, 

i.e. according as 

tan ^ - - < 0, = 0, or > 0. 
a 

But h/a is the tangent of the angle <^ which an element of the 
cone with vertex S makes with the plane a. If is less than 
this angle (f>, the section of the cone is an ellipse ; if ^ is equal 
to <^, the section is a parabola ; and if 6 is greater than <^, 
the section is a hyperbola. Note that this result is in accord- 
ance with our geometric intuition of the situation. 



XIII, § 242] THE PARABOLA 373 

EXERCISES 

1. Prove that the central projection of any circle is a conic ; that is, 
that a plane section (not through the vertex) of any circular cone (not 
necessarily a right cone) is a conic. 

[Hint : Complete generality will be secured by taking the equation of 
the circle in a to be x'^ ^- y'^ -{■ dx -\- c = 0. Why ?] 

2. Prove that the central projection of any conic is a conic. 

,3. Prove that the central projection of a straight line is a straight line. 

4. Prove that there exists in a just one straight line which has no 
corresponding line in a', namely the line of intersection of a with the 
plane through S parallel to a'. This line is called the vanishing line of a. 

5. Prove that the central projection of a circle in a is an ellipse, a 
parabola, or a hyperbola, according as the vanishing line in a meets the 
circle in no points, one point, or two points. 

242. Poles and Polars. Diameters. We have had occasion 
in several exercises to note that the equation which represents 
the tangent to a conic at the point Piix^ , y^ when Pj is on the 
curve, represents a straight line called the polar of Pi when Pj 
is any point in the plane. Pi is then called the pole of the 
line with respect to the conic. The polar of a point on the 
conic is then the tangent at the point. We have also seen 
that the polar of a point Pi through which pass two tangents 
to the conic is the line joining the two points of contact of the 
tangents. In the more extensive geometric theory of conies 
poles and polars play an important role. 

A straight line passing through the center of an ellipse or 
hyperbola is called a diameter of the conic. Every diameter 
of an ellipse meets the curve in two points ; some of the 
diameters of a hyperbola meet the curve in two points. These 
points are then called the extremities of the diameter, and the 
distance between them is called the length of the diameter. 
Any line parallel to the axis of a parabola is called a diameter 
of the parabola. Other properties are given in exercises below. 



374 MATHEMATICAL ANALYSIS [XIII, § 242 

MISCELLANEOUS EXERCISES 
Properties of Poles and Polars 

1. Write the equation of the polar of each of the following points 
with respect to the conic given, and draw the corresponding figure : 

(a) (1,5); 2a;2 + y2^4. (d) (- 1, 3) ; a;2 + 2/2 + 4 x - 6 y- 2 = 0. 

(6) (2,0); 4x2-92/2 = 36. (c) (2, - 3) ; 2/2 = 6x. 

2. Find the pole of the line 3x — 4y4-12 = with respect to tho 
following conies : 

(a) 3x2 + 4 y2 = 12 ; (ft) a;2 - 5 y2 = 20 ; (c) y^ = Sx; (d) x'^ = iy. 

3. Prove that in any conic the polar of a focus is the corresponding 
directrix. 

4. Prove that in any conic, if Pi and P2 are two points such that the 
polar of Pi passes through P2, the polar of P2 will pass through Pi. 

5. From the result of the last exercise follows geometrically the follow- 
ing theorem : If a straight line be revolved about a point P and tangents 
are drawn at the points where it meets a conic, the locus of the intersec- 
tion of these pairs of tangents is the polar of P with respect to the conic, 
or a part of the polar. Which part will it be ? 

6. Prove that the polar of any point on a directrix of a conic passes 
through the coiTesponding focus. [See Exs. 3 and 4.] 

7. A straight line through a point Pi meets a conic in Ci and C2, and 
the polar of Pi in Q. Prove that Pi and Q divide the segment C1C2 in- 
ternally and externally in the same ratio. 

[Solution : Let Pi(xi, yi) and P2(X2, 2/2) he any two points. Then 
the point P whose simple ratio with respect to Pi and P2 is X has the co- 
ordinates ^ _ xi + Xx2 , _ 2/1 + X1/2 

If these be substituted in the equation of the ellipse bH^ 4- a^y^ = ajb^ and 
the resulting equation arranged as a quadratic in X, we have 

/^4.y22_l^^2 + 2f^2^M_2_l^x + f^' + ^'-lUo. 

V a2 ^ 62 ) ^ \ a^ ^ b^ J \a^ b^ J 

The roots of this equation are the simple ratios of the points Pi and P2, 
respectively, with respect to the points d, C2 in which the line P1P2 meets 
the ellipse. If the segment Ci, C2 is to be divided internally and ex- 
ternally in the same ratio the roots Xi, X2 of this equation must be equal 
numerically, but opposite in sign, i.e. Xi -f- Xo must be zero. The coefficient 



XIII, § 242] THE PARABOLA 375 

of X in the above equation would then be zero, if the theorem to be proved 
is true. But the condition 

«2 ^ 62 

is precisely the condition that P'z{x2^ y-2) be on the polar, 

of Pi with respect to the ellipse. The similar proofs for the hyperbola and 
parabola are left as exercises.] 

8. Two points P, Q on the line joining two given points d, C2, are 
said to divide the segment Gi C-i, harmonically, if they divide the seg- 
ment internally and externally in the same ratio (i.e. if CxP/PCi = 
— O1Q/QC2). Show that the result of the last exercise leads to the fol- 
lowing : The locus of a point Q^ such that a given point P and the point 
Q divide harmonically the segment joining the points in which the line PQ 
meets a conic is the polar of P with respect to the conic, or a part of the 
polar. Which part is it ? (Compare with the similar question in Ex. 5.) 

Properties of Diameters 

9. ProTB that the locus of the mid-points of the chords of a conic 
drawn parallel to a given chord is a diameter of the conic. 

[Solution for the Ellipse : Let the equation of the ellipse be 

b^2c^ + a^y^ = a^b'^, and let the slope of the given chord be m. Then any 

chord parallel to the given chord isy = mx + k. The abscissas Xi, X2 of 

the points in which this chord meets the ellipse are the roots of the 

equation 

(62 + a2^2)a;2 + 2 a^mkx + a^(k^ - 62) = 0. 

The sum of the roots of this equation is 

Xi-}-X2= — 



62 -I- a^n^^ 



The coordinates (x', y') of the mid-point of the chord are then 

X' = itixi -f X2) = - TT^^, y' = mx' +k= ^'^ 



62-|-a2m2' 62 + a2m2 

The coordinates x', y' then satisfy the equation y = — (b^x^/(a^m), no 
matter what the value of k is. The locus of the mid-points of the 
chords whose slope is m is, therefore, the straight line whose equation is 
y = — (62x)/(a2m). Since this straight line passes through the center of 



376 MATHEMATICAL ANALYSIS [XIII, § 242 

the ellipse, the theorem is proved for the ellipse. The similar proofs for 
the hyperbola and parabola are left -as exercises.] 

10. From the result of the last exercise, show how to construct a 
diameter of any conic, and hence (in case of ellipse and hyperbola) how 
to find the center, when only the outline of the curve is given. 

11. Having given the outline of an ellipse or hyperbola, construct the 
center. Then show how to construct the principal axes (make use of the 
fact that the principal axes are axes of symmetry ; a circle drawn with 
the center of the conic as center and suitable radius will meet the conic in 
the four vertices of a rectangle whose sides are parallel to the principal 
axes). Then construct the foci and the directrices. 

12. Having given only the outline of a parabola, show how to construct 
the axis, the focus and the directrix. 

13. Show that the tangent drawn to a conic at an extremity of a 
diameter is parallel to the chords which the diameter bisects. 

14. If two diameters of a conic are such that each bisects the chords 
parallel to the other, the diameters are said to be conjugate; and each 
is called the conjugate of the other. Prove that if wi, W2 are the slopes 
of two conjugate diameters of the ellipse h'^x'^ + a'^y'^ = d^h'^, then we have 
m\m2. = — ly^/a^. 

15. Prove that, if wi, mi are the slopes of two conjugate diameters 
of the hyperbola fe^x^ _ a'iyi — a-h'^, we have WiW2 = h'^/a^. 

16. The only conic for which all pairs of conjugate diameters are per- 
pendicular is the circle. 

17. The polars of the points on any diameter of an ellipse or hyperbola 
are parallel to the conjugate diameter. 

18. If one extremity of a diameter of an ellipse Jfix^ -\- a^y'^ = aV)^ 
has the coordinates (xi, yi), one extremity of the diameter conjugate to 
the given one will have the coordinates (— yia/b, X\b/a). 

19. The area of a parallelogram circumscribed about an ellipse whose 
sides are parallel to two conjugate diameters is constant and equal to 4 db. 

20. Prove that, if a\ and &i are the lengths of two conjugate semi- 
diameters of an ellipse, a\^ + fti^ = a^ -f h^. 

21. Prove that any pair of conjugate diameters of the hyperbola 
6%2 _ 052^2 _ 0,252 are also conjugate diameters of h'^x^ — a^y'^ = — aP'h'^. 

22. If a diameter of a hyperbola with center meets the hyperbola in 
P and the conjugate diameter meets the conjugate hyperbola in Q, prove 
that 0P2 _ 0q^ = (£^ - 62. 



CHAPTER XIV 
POLAR COORDINATES 

243. Review. Polar coordinates, introduced in §§ 112-114, 

are often useful in studying geometry analytically. The 
present chapter is devoted to illustrating some of the principles 
involved and their applications. 

EXERCISES 

1. What is the locus of points for which p is constant ? 

2. What is the locus of points for which 6 is constant ? 

3. Show that the points (p, d) and (/s, — 0) are symmetric with respect 
to the polar axis. 

4. Show that the points (/>, 6) and (— p, 6) are symmetric with respect 
to the pole. 

6. Show that the points {p,.d) and (/>, d + 180°) are symmetric with 
respect to the pole. 

6. Find the distance between the points A (2, 45 '^) and B(7, 105°). 
[Hint. Use the law of cosines.] 

7. Prove that the distance between the points (pi, ^i) and (p2, ^2) is 



Vpi'^ + pi^ - 2 P1P2 cos {62 — di). 



244. Locus of an Equation. The locus of an equation in the 
variables p and is such that : 

(1) Every point whose coordinates (p, 0) satisfy the equation 
is on the locus or curve, and 

(2) A set of coordinates * of every point on the locus or curve 
satisfies the equation. 

* Not necessarily every set. Thus, the point (2, 60°) = ( - 2, 240°) is on the 
locus of p = 1 + 2 cos 6 ; but the second set of coordinates does not satisfy the 
equation. 

377 



378 



MATHEMATICAL ANALYSIS [XIV, § 244 



The curve may be sketched by computing a table of corre- 
sponding values of p and 6, plotting the corresponding points, 
and then sketching the curve through them. The amount of 
work may often be shortened if one makes use of the following 
obvious rules for symmetry : 

If a polar equation is left unchanged, 

(a) ichen 6 is replaced by — 6, the locus is symmetric with re- 
spect to the polar axis. 

(b) when p is replaced by — p, the locus is symmetric with re- 
spect to the pole. 

(c) ivheyi 6 is replaced by 180° + 6, the locus is symmetric with 
respect to the j)ole. 

(d) ivhen 6 is replaced 6?/ 180° — 9, the locus is symmetric ivitJi 
respect to the line through the pole perpendicular to the polar axis. 

It should be borne in mind, however, that none of these rules 

are necessary conditions for sym- 
metry. Why not ? 

245. Illustrative Examples. 

We shall illustrate the methods 
of plotting curves in polar coordi- 
nates by the following examples. 

Example 1. Discuss and plot the 
locus of the equation p = 4 cos 6. 

The locus is symmetric with respect 
to the polar axis. If we plot points 
from 0° to 90°, we obtain the upper 
half of Fig. 215. Then by symmetry 
we obtain the complete graph given in 
Fig. 215. Why ? 



Fig. 215 




d 


0° 


30° 


45° 


60° 


90° 


p 


4 


3.5 


2.8 


2 






XIV, § 245] 



POLAR COORDINATES 



379 



Example 2. Discuss 
and plot the locus of the 
equation p — 4t &m'^ 6. 

The locus is symmetric 
with respect to the pole, 
the polar axis, and the Ime 
through the pole perpendic- 
ular to the polar axis. If 
we plot points in the range 
^=0° to 0=90°, and make 
use of symmetry, we have 
the complete figure which 
is given in Fig. 216. 



e 
p 


0° 



30° 

1 


45° 
2 


C0° 
3 


90° 
4 




Fig. 216 

The branches constructed 
by symmetry should be 
checked by substituting in 
the original equation the 
coordinates of at least one 
point on each branch. Seri- 
ous errors may thus be 
avoided. 

Example 3. Discuss and 
plot the locus of the equa- 
tion p =: 1 -h 2 cos (9. 

The curve is symmetric 
with respect to the polar 
axis. If we plot points 
from 6 = 0° to 6= 180°, we 
get the points shown in 

Fig. 217. Then by symmetry we get the complete graph, or the curve 

in Fig. 217. 




Fig. 217 



6 


0° 


30° 45° 


60° 


90° 


120° 


135° 150° 


180° 


P 


3 


14-V3 


1 -H\/2 


2 


1 





1-V2 


1-V3 


-1 



380 MATHEMATICAL ANALYSIS [XIV, § 245 

EXERCISES 

Are the following loci symmetric with respect to the pole ? The polar 
axis ? The line through the pole perpendicular to the polar axis ? 



1. p = a cos 0. 




5. 


p- = a cos 2 d. 


9. 


p = a sin2 6. 


2. p = a sin d. 




6. 


p'^ = a sin 2 6. 


10. 


P = sin^l 
p = e. 


3. p = a (1 — cos 


d). 


7. 


p = a cos 2 ^. 


11. 


4. p = a(l — sini 


9). 


8. 


p = a sin 2 ^. 


12. 


p2 cos 0=4. 


Discuss and plot 


, thel 


ocus 


of each of the following < 


equations. 


13. p = 5. 




22. 


p cos ^ = 4. 


31. 


p =z 1 + 2 sin e. 


14. p=- 5. 




23. 


p cos ^ = — 4. 


32. 


p = 1 — 2 cos d. 


15. p2 = 25. 




24. 


p sin ^ = 5. 


33. 


P = 1 — 2 sin 0. 


16. ^ = 30°. 




25. 


p sin ^ = - 6. 


34. 


p = 2 + cos 0. 


17. e=- 30°. 




26. 


p = 1 - cos ^. 


35. 


p = 2 + sin d. 


18. p = 8 cos d. 




27. 


p = 1 + cos ^. 


36. 


p = 4 tan d. 


19. p = - 8 cos e. 




28. 


p = 1 ~ sin ^. 




1 






29. 


p = 1 + sin 0. 


37. 


p — '■ 


20. p = 8 sin e. 


l-cos^* 


21. p = — 8 sin e. 




30. 


p = 1 + 3 cos ^. 







246. Standard Equations. We shall now derive polar equa- 
tions for the straight line and the conic 
sections. 

The straight line. Let CD be any straight 
line (Fig. 218) ON = p the perpendicular 
let fall upon it from the pole 0, and a the 
angle which this perpendicular makes with 
the polar axis. Let (p, 0) be any point on 
the line. 

ON 
Then --- = cos (0 — a) or cos (a — 6). 

But by § 120, 

cos (0 — a) = cos (a — $). 
Hence 

(1) P = p cos (6 - a) 

is the desired equation. 




XIV, § 246] POLAR COORDINATES ' 381 

If the line is perpendicular to the polar axis, its equation is 
p cos ^ = p. Why ? 

If the line is parallel to the polar axis, its equation is 
p sin ^ = p. Why ? 

The circle. Let C (c, a) be the center of a circle of radius r 
and P (pj 6) any point on the curve (Fig. 219). In the triangle 




Fig. 219 



OOP, OC=c, OP = p and the angle COP = ± (0 - a) 
depending upon the position of the point P. But since 
cos (0 — a) = cos {a — 0), we have from the law of cosines, 
§126, 

(2) r2 = c2 + p2 _ 2 c p cos (6 - a) 

as the equation of the desired locus. 

If the center C is upon the polar axis (a — nir), equation (2) 
becomes 

(3) r2 = c2 -f p2 ± 2 c p cos d. 

If the circle passes through the pole (r = ± c), equation (2) 
becomes 

(4) p = ± 2rcos(^-a). 

If the center C is upon the polar axis (a = 0) and the circle 
passes through the pole (c = ± r), equation (2) becomes 

(5) p = ± 2 r cos 6. 

If the center is at the pole (c=0), equation (2) becomes p= ±r. 



382 



MATHEMATICAL ANALYSIS [XIV, § 246 



The Polar Equation of any Conic. The polar equation of 
any conic may now be derived. Let DD' be the directrix, F 

the corresponding focus, and e 
the eccentricity. Let the per- 
pendicular through F to DD' 
meet DD' in L. Let the segment 
LF = p, and take F as the pole 
and the extension of the line LF 
as the polar axis. If P(p, 6) is 
any point on the curve and PS 
is the perpendicular from P to 
DD\ then by the definition of a 

PS, 




Fig. 220 



conic, we have 
that is, 



or 



(6) 



FP==e 

p = e{p + p cos 6), 



1 — e cos 6 



which is the polar equation of a conic. If e < 1, the equation 
represents an ellipse ; if e = 1, a parabola ; e > 1, a hyperbola. 



EXERCISES 

1. Derive the equation p = 2 r cos [ (5) , § 246] directly from a figure. 

2. Derive tiie polar equation of the ellipse assuming the right-hand 
focus as the pole and the major axis as the polar axis. 

3. Derive the polar equation of a hyperbola assuming the right-hand 
focus as the pole and the transverse axis as the polar axis. 

4. Derive the polar equation of the circle which passes through (0, 0°) 
and has its center at (r, 90°) ; (r, 270^). 

6. Derive the polar equation of the parabola assuming the focus at the 
pole and the directrix the line p sind = p ; psind = — p. 

6. The difference of the focal radii of a certain hyperbola is 3, and the 
distance between the foci is 6. Find a polar equation of the curve. 



XIV, § 247] 



POLAR COORDINATES 



383 



247. Other Curves. What is the advantage of polar coordi- 
nates ? Why not continue to use only rectangular coordinates ? 
The answer to these questions is that in certain kinds of 
problems polar coordinates are much more convenient. The 
following examples will illustrate the desirability of polar 
coordinates. 

The limacon. Through a fixed point upon any given circle 
of radius a, a chord OPi is drawn and produced to P so that 




Fig. 221 

P^P = kj where A; is a given constant (Fig. 221). Find the 
locus of P"* as Pi describes the circle. 

If p — 2 a GosO is the 
equation of the circle and 
the pole is the fixed point, 
then the locus of P is 

(7) p = 2 a cos 0-\-Jc. 

If k = 2 a, the equation 
may be written in the form 

(8) p = 2 a(l + cos 0) 

and the curve is known as the 
cardioid, on account of its 
heart-shaped form (Fig. 222). fig. 222. - The Lima9on 

* This curve is known as the lima<^on of Pascal. It was invented by 
Blaise Pascal (1623-1662) , a famous French mathematician and philosopher. 
The word lima(;on means snail. The Germans call this curve die PascaVsche 
Schnecke. 




384 



MATHEMATICAL ANALYSIS [XIV, § 247 



The cissoid. OA is a fixed diameter of a fixed circle (Fig. 
223). At the point A a tangent is drawn, while about the 
point a secant revolves which meets the tangent in B, and 
the circle in C. Find the locus of a point P on OB so deter- 
mined that 0P= CB. 




Fig. 223 



Take as the pole and OA as the polar axis of a system of 
polar coordinates. If we denote OA by 2 a, then 
the equation of the circle is p= 2 a cos 6. Let P 
be denoted by (p, 6). Now 

p=OP=OB-PB. 

0B=2aseGe and PB = 0C=2acose. 

p = 2 a (sec — cos 0), 



p = 2 a tan ^ sin 0. 
Fig. 224. — The 

Cissoid The locus of this equation is given in Fig. 224. 

The curve is known as the cissoid of Diodes.* 




* Cissoid (Greek, Kto-o-ds = ivy) means ivy-like. The Greeks considered 
only the part of the curve lying within the circle. Diocles was a Greek 
mathematician who lived sometime between 217 b.c. and 70 b.c. By means 
of this curve, Diocles showed how to construct the side of a cube whose 
volume is twice the volume of a given cube. See Ex. 4, p. 388. 



XIV, § 247] 



POLAR COORDINATES 



385 



Conchoid of Nicomedes.* A straight line revolves about a 
fixed point and meets a fixed straight line MN in the point 
Q. From Q a fixed length is laid off on OQ in both direc- 
tions. The locus of the two points, P and P', thus determined 
is called a conchoid. 

Let be the pole and the line OR through perpendicular 
to MJSf be the polar axis of a system of polar coordinates 




Fig. 225. —The Conchoid 

Let (p, 6) be the coordinates of any position of the generating 
point P (or P'). Then 

p = OP(or OP') = OQ±QP=OBseGe± QP. 

But OR and QP are given constants ; call them a and b 
respectively. Then 

(10) p = asec^±6 

is the desired equation of the conchoid. 

* Conchoid (Greek, Koyxo? = mussel) means mussel-like. Nicomedes was a 
contemporary of Diodes. He invented the conchoid for the purpose of trisect- 
ing an angle, which is one of the famous problems of antiquity. This prob- 
lem cannot be solved by means of the compass and straightedge alone. 
2c 



386 



MATHEMATICAL ANALYSIS [XIV, § 248 



248. Spiral Curves. A spiral is a curve traced by a point 
which revolves about a fixed point called the center, but con- 
tinually recedes from or continually approaches the center ac- 
cording to some definite law. 

The spiral of Archimedes is the locus of a point such that its 
radius vector is proportional to its vectorial angle. Therefore 
its equation is 

(11) p = ke, 

where Ic is a constant.* 

The form of the equation shows that the locus passes through 
the pole, and that the radius vector increases without limit as 




the number of revolutions increases without limit. Figure 226 
represents a portion of the locus for k = ^j with $ expressed 
in degrees. 

The hyberbolic or reciprocal spiral is the locus of a point 
such that its radius vector is inversely proportional to its vec- 



* In this example, and in those that follow, it is usual to express the angle 
in radians ; but this is not necessary, since the same result can be obtained by 
choosing a different value for A: if d is exi>ressed in degrees. 



XIV, § 248] 



POLAR COORDINATES 



387 



torial angle. The equation of the locus is therefore 

(12) P=^. 
where A; is a constant. 
Figure 227 represents a 
portion of the graph for 
A; == 70 and for positive 
values of 6 (expressed in 
degrees). 

The logarithmic spiral 
is the locus of a point ^^^- ^27 

such that the logarithm of its radius vector is proportional to 
its vectorial angle, i.e. 

(13) log p = kO, 





where A; is a constant. If the base of the system of logarithms 
is b, the equation may be written in the form p = 6*^. Figure 
228 represents a portion of this locus when 6 = 3, for A: = y^, 
with 6 expressed in degrees. 



388 



MATHEMATICAL ANALYSIS [XIV, § 248 



EXERCISES 

1. Discuss the form of the limagon (7), when | A: | < 2 a. When 
|A:|>2a. 

2. Solve the locus problem used to define the lima9on by means of rec- 
tangular coordinates, and compare the merits of the two solutions. 

3. By taking the line OA (Fig. 223) as the ic-axis and the tangent to 
the circle at O as the y-axis, prove that the equation of the cissoid is 

2a-x 

4. Duplication of a Cube. In the adjoining figure, let MN= a and 

MB = 2 a. Draw BA and let it meet the cissoid 
in the point D whose ordinate is LD. Prove 
that ZZ)3 = 2 OLK If MB = n • a prove that 
LD^ = n ■ OL^. 

6. If in Fig. 225 the line MN^ is taken as the 
a;-axis and the line OB A as the jz-axis, prove that 
the equation of the conchoid is 

a:V=(y + «)'(&' -2/2). 
Compare the merits of this solution with that on 
p. 385. 

6. Trisection of an angle. Let AOB be the angle to be trisected. 
Through a convenient point A on one side OA of the angle draw AB per- 
pendicular to OA. Through 
B draw a line B C parallel to 
OA. From as fixed point, 
and AB as fixed line, and 
2 . OB as a constant dis- 
tance, describe a conchoid 
meeting BC in C. Angle 
^OC is then \ AOB. 

[Hint: E is the mid-point 
of Z>C ; then OB = BE = EC. The result then follows from elementary 
geometry.] 

7. Show that in the conchoid, if 

(a) 6 > a, the curve has an oval at the left, as in Fig. 226; 
(6) 6 = a, the oval closes up to a point ; 

(c) 6 < a, there is no oval and both branches lie to the right of the 
point 0. 





XIV, § 249] 



POLAR COORDINATES 



389 



8. Draw the parabolic spiral, which is defined by the equation p^ = kd. 
Take k = -^^ with 6 in degrees, and use only the positive values of p. 

9. Draw the lituus, which is defined by the equation p^ = k/d. Take 
A; = 90 with 6 in degrees, and use only the positive values of p. 

249. Relation between Rectangular and Polar Coor- 
dinates. Take the origin of a system of rectangular axes 
as the pole, and the positive half of the a;-axis as the polar 
axis, of a system of polar coordinates. 

Let (x, y) and (p, 6) be respectively the rectangidar and polar 
coordinates of any point P. Then x/p — cos ^, yjp — sin 6. 
Hence, we have 

I X = p cos e, 

I y = p sin e. 



(14) 



It is here assumed that the coordinates of P are so chosen that 
OP — p and angle XOP = 0. This is always possible. If p is 
positive, X always has the sign of cos and y the sign of sin $, 




ft<=^' -?^ 



F P 

Fig. 229 

Conversely, if p is positive, we see from Fig. 229 that 
.2 _ ^2 _L „2 gin $ — ^ 



(16) 



x" + y^, 

$ = arc tan [t\ cos d = ■'- ' 



390 MATHEMATICAL ANALYSIS [XIV, § 249 

EXERCISES 

Transform the following equations into equations in rectangular coordi- 
nates. State in each case whether the graph is easier to sketch from the 
polar or the rectangular equation. 

1. p = 1 - cos 0. 6. p2 sin 2 e = 3. 11. p^ = 6. 

2. p=l + sin^. 7. p2cos2^ = 4. ^^ ^2^1 



3. p = 4. 



8. p2 = cos 2 e. 



13. p = a sec d-\- b. 



4. pcos^ = 5. I 14. p = 2asec^tan^. 

10. p = — • 

6. p sin ^ = - 2. e 15. p = 4 cos 2 6. 

Transform the following equations into equations in polar coordinates : 

16. ^2 + 1/2 = 4 X. 21. xy = 4. 

17. (x2 + y2)2 _aj2 _ 2/2. 22. a; cos a 4- y sin a = ;>. 

18. X -y = 0. 23. (y2 4- x2 _ 2 a;)2 = x2 -f- y2. 

19. 1/2 = 4 X. 24. x3 = ?/2(2_a;). 

20. 9 x2 + 4 ?/2 = 36. 25. x2?/2 = (y + 2)2(4 - y^). 





MISCELLANEOUS 


EXERCISES 


Sketch the follow 


fving 


curves : 




1. p = a cos 2 d. 






1. p = a cos 6 6. 


2. p = a cos 8 d. 






8. p = a sin 5 ^. 


3. p = a sin 2 ^. 

4. p = a sin 3 6. 






• 6 
9 p = a sin-- 


h. p = a cos 4 ^. 
6. p = a sin 4 ^. 






10. p = acos-- 



Find the points of intersection of the following pairs of curves. Plot 

the curves in each case and mark with their respective coordinates the 
points of intersection. 

11. p == 1 4- cos e,. 14. p = 1 + cos d, 
4(1 +cos0)p = 1. 2p = 3. 

12. p = 4, 15. p = 2(l -sin^), 
pcos^=2. (l+sin^)p=l. 

13. p = \/2, ^®- P = cosd, 

P = 2 sin 5. P = 1+ 2 cos d. 



XIV, § 249] POLAR COORDINATES 391 

Solve the following exercises by the use of polar coordinates : 

17. Find the locus of the center of a circle which passes through a 
fixed point O and has a radius 2. 

18. Prove that if from any point a secant is drawn cutting a circle in 
the points P and Q, then OP - OQ is constant for all positions of the 
secant. 

[Hint. By using equation (2), §246, show that the product of the 
roots is constant.] 

19. Secants are drawn to a circle through a fixed point on the cir- 
cumference. Find the locus of the middle points of their chord segments. 

20. Find the locus of the middle points of the focal radii issuing from 
one focus of an ellipse ; parabola ; hyperbola. 

21. The focal radii of a parabola are produced a constant length. Find 
the locus of their end-points. 

22. Through a fixed point on a fixed circle a variable secant OP is 
drawn cutting the circle in B. If BP = 3 OB, find the locus of P. 



CHAPTER XV 

PARAMETRIC EQUATIONS 

250. Parametric Equations. As a point P(xj y) moves along 
a given curve, the x- and y- coordinates of the point vary. So 
do many other quantities connected with this point, as for ex- 
ample, in general, its distance OP from the origin, the angle 
which OP makes with the aj-axis, its distance from a fixed line, 
etc. It is sometimes convenient to express x and y in terms of 
one of these variables. This third variable, in terms of which 
the variables x and y are expressed, is called a parameter. For 
exaonple, we see that the coordinates of any point P{x, y) on 
the circle whose center is at the origin and whose radius is r, 
can be expressed in the form 



(1) rx = rco 

I 1/ = r si 



r cos 0, 
sin 6, 




Fig. 230 

where 6 is the angle XOP (Fig. 230). These are then para- 
metric equations of the circle. If we eliminate the parameter 
6 between these two equations by squaring and adding them, 

we obtain the equation 2 _j_ ,,2 _ ^2 
X -i- y — r , 

which is the rectangular equation of the circle. 

392 



XV, § 250] PARAMETRIC EQUATIONS 

Similarly, a pair of parametric equations of the ellipse 

(2) 
are 
(3) 



393 



^4-^ = 1 
a2 62 



f X = a cos 6, 
1 y = 6 sin 6 ; 

for, these values of x and y are seen to satisfy equation (2) 
for all values of 6. 

The geometric interpretation of equations (3) is important. 
In Fig. 231, a geometric construction given in § 226 is used. 




Fig. 231 

The abscissa of P is equal to the abscissa of Q, i.e. a cos 0, the 
ordinate of P is equal to the ordinate of R, i.e. b sin 6. There- 
fore the coordinates of P are x = a cos 0, y = b sin 0. The 
angle is known as the eccentric angle of the point P. We 
should note that is not the angle XOP. 

A pair of parametric equations for the hyperbola 

(4) ^^-yl=i 

^ ^ a^ b^ 

are 

(5) x = a sec 9, y = b tan 9, 

for these values of x and y satisfy the equation (4). 

It is important to note that a given curve may have as many 
sets of parametric equations as we please. For example, para- 



394 MATHEMATICAL ANALYSIS [XV, § 250 

metric equations of a circle may be written in the form 
x = a cos t, y = a sin t, 

as above ; or they may be written in the form 

or in any one of many other forms. 

EXERCISES 

1. Show that x = t, y =2 — t are parametric equations of a straight 
line. 

2. Show that x = I pt^, y =pt are a pair of parametric equations of 
the parabola y'^ = 2 px. 

3. Write two pairs of parametric equations for the line y = x. 

4. Prove that ., ,„. „ , 

are parametric equations of a circle. 

6. Write a pair of parametric equations for the rectangular hyperbola 

a;2 - ?/2 _ ^2, 

6. Show that x = Acosd -\- Bsin d, y = Asm 6 — Bcosd are para- 
metric equations of a circle. 

7. Prove that x = 6 + 4 cos ^, y = — 2 + 3 sin ^ are parametric equa- 
tions of an ellipse. 

8. Write a pair of parametric equations for the circle 

(x - a)2 + (y - 6)2 = r2. 

9. Prove that a; = 6 + 4sec^, y=— 2 + 3 tan d are parametric equa- 
tions of a hyperbola. 

10. Find the equation of the tangent to 

X^ f/2 

(<^) -^ + 7^ = ^» *^ ^1 = « <^08 ^i» yi = ft sin ^1- 

/v2 «f2 

(ft) — o — t:; = 1' at ^1 = « sec 01, yi = 6 tan ^i. 
(c) ?/2 = 2px, at xi = i ;)«i2, yi = pti. 



XV, § 251] 



PARAMETRIC EQUATIONS 



395 



11. Prove that the tangents to y''- = 2px at {^ph^.ph)^ ilPh^.Ph) 
meet at the point [^^phti, \p{h + ^2)]- 



12. Write the equation of the tangent to y^= 4 ax at xi = 

13. Show that 



Wli^' 



2a_ 



3a« 



y = 



3a«2 



1 + «3 ' ^ 1+^3 

are parametric equations of the curve x^ -{-y^ —Z axy. 

14. Show that x = a cos^ e, y = a sin^ d are parametric equations of 

21 2 
the curve x'^ -i- y^ = a^. 

15. Find the x- and ^/-equations of the curve whose parametric equa- 
tions are X = a (^ — sin d), y = a (1 — cos d). 

251. Sketching Loci of Parametric Equations. If we assign 
a series of values to the parameter and determine the series of 
corresponding pairs of values for x and 2/j we can interpret 
these values as the coordinates of points on a curve. Plotting 
these points and sketching a curve through them, we have the 
graph of the curve whose parametric equations were given. 

Example. A pair of parametric equations giving the path of a 
body projected horizontally from a height of 400 ft. with a velocity of 
10 ft./sec., are x=10t, y = 400 — 16 t^. Sketch the locus. 



t 





1 


2 


3 


4 


5 


X 





10 


20 


30 


40 


50 


y 


400 


884 


336 


256 


144 






Y 

400 
300 
800 
100 


















N 
















\ 


\ 














\ 


V 














\ 









1 


a 


s 


4 


6 


6 


X 



Fig. 232 

In the preceding table are given the values of x and y corresponding 
to the integral values of t from to 5 inclusive. Plotting these points 
we have the graph in Fig. 232. 

This curve is of course the same that we should obtain by first elimi- 
nating t and then plotting from the equation in x and y. 



396 MATHEMATICAL ANALYSIS [XV, § 252 

252. The Time as Parameter. Suppose a point moves in 
a plane. At every instant of time t the point occupies a certain 
position (x, y). In other words, the coordinates x and y of the 
point P are functions of t, i.e. 

,/,x r a; = a function of ty 

I y = a function of t. 

These equations are then parametric equations of the path 
traversed by the point. 

Such equations arise frequently in mechanics when it is 
desired to describe the motion of a body subject to various 
forces. For example, if a body is projected from a point 
(0, 0) in a vertical plane at time i = 0, with an initial velocity 
Vq, and making an angle a with the horizontal (ic-axis), its 
position at the end of t seconds * is given by the equations 

{x = tv^ cos a, 

[y = tVoSma-}gt% 

where, if Vq is measured in ft./sec, gr is a constant approxi- 
mately equal to 32.2. The use of these equations of a projectile 
will be illustrated in the next article and the exercises follow- 
ing it. 

EXERCISES 
Sketch the following curves from their parametric equations. 



1. x = t, 


4. X = t'^ - 1, 


7. x = t, 


lo. 


x = t, 


2/ = « + 2. 
2. x = r^, 


y = t^-\. 

5. X = ^2 4. 1, 


1 


11. 


y = t-tK 
X = sin d, 


y = r. 


y = z. 


8. x = t\ 




y = cos d. 


3. x = s+ 1, 


6. x = t, 


y = t\ 


12. 


x = tan e, 


y = s'^. 


y-\- 


9. x = t^+\, 
y = t^^l. 




y = sec d. 


13. x = \Qt cos SO'', 


14. x = 5«, 






?/ = 25 + < 


sin 30° - 16 «2. 


y = so- 


16 «2. 





* The resistance of the air being neglected. 



XV, § 253] PARAMETRIC EQUATIONS 



397 



253. The Path of a Projectile in Vacuo. The equations (7) 

given in § 252 yield many results of interest regarding the paths of pro- 
jectiles. Some of these are given in this article and others are found in 
the exercises below. They are, of course, only approximations to the 
actual behavior of a projectile, in view of the fact that the resistance of 
the air has been neglected * 

By eliminating t between the equations (7), § 252, we obtain the equa- 
tion of the path in rectangular coordinates (x, y) : 



(8) 



y = X tan a — 



gx^ 




Fig. 233 

The path is, therefore, a parabola, with a vertical axis. The vertex 
of the parabola is at the point (Fig. 233) 



(9) 






\m 2 a, <''''''' ' 



^9 



The greatest height above the horizontal is 



(10) 



H = 






The complete range, i.e. the distance from O to the point where the 
projectile again meets the horizontal, is found as follows : 

If in (7) we place y = 0, we find t = and 
t = 2{vjfj) sin a. 
The value of x for the second value of t found is the desired range B, i.e. 

.2 



B 



^ sin 2 a. 

g 

This result could also have been found by placing y = in (8) and solv- 
ing for X. Why ? 

* For the theoretical and practical discussion of the flight of actual projec- 
tiles (whose motion is 'appreciably affected by the resistance of the air) the 
student is referred to Alger, The Groundwork of Naval Gunnery, or 
External Ballistics. 



398 MATHEMATICAL ANALYSIS [XV, § 253 

EXERCISES 

1. A gun is fired at an elevation of SO'^. Find the range if the muzzle 
velocity of the shell is 1000 ft./sec. Aiis. About 5 mi. 

2. What is the greatest height reached by the projectile in Ex. 1 ? 
How long is its time of flight ? 

3. What must be the initial velocity of a baseball thrown at an angle 
of 45° in order that it may travel 200 ft. before hitting the ground ? 

4. A stone is thrown from a tower 100 ft. high, with an angular ele- 
vation of 45° and an initial velocity of 64 ft./sec. How far from the foot 
of the tower will the stone hit the ground ? 

6. The great pyramid of Cheops is 450 ft. high. Its base is a square 
746 ft. on a side. A ball is thrown upwards from the top in a direction 
making an angle of 20° with the horizontal and with the velocity of 
80 ft./sec. Will the ball clear the base of the pyramid ? 

6. Prove that for a given initial angle of elevation the range of a pro- 
jectile is proportional to the square of the initial velocity. 

7. Prove that for a given initial velocity the maximum range is 
obtained when the angle of elevation is 45°. 

8. Prove that with the notation of § 253, the time of flight of a pro- 
jectile from to (x, y) is (x/vo) sec a ; from Oto (J?, 0) is (2 Vo/g) sin a. 

9. Prove that the paths of a projectile with given vq, but varying a, 
have the same directrix. 

10. Prove that the coordinates of the focus of the path of a projectile 

are 

( vq^ sin 2 a — ih)^ cos 2 a \ 
2sr ' 2g )' 

Hence show that the locus of the foci of all paths in a given vertical plane 
with the same vo is a circle with center at 0. 

11. Prove that the parabola of maximum range has its focus on the 
avaxis. 

12. Prove that the locus of the vertices of the paths with given "yo is 
an arc of an ellipse. 

13. Prove that the locus of the vertices of the paths with a given a. 
and a varying vo is a straight line. 

14. Prove that the locus of the foci of the paths with a given a and a 
varying vo is a straight line. 



XV, § 254j 



PARAMETRIC EQUATIONS 



399 



H 



^ 



254. Locus Problems. Parametric equations of a curve 
are sometimes much more easily obtained and easier to work 
with than either the rectangular or polar equations. The 
following problems illustrate some 
of the methods that may be em- 
ployed. 

Example 1. A line of fixed length 
moves so that its ends always remain on 
tlie coordinate axes. Find the locus gen- 
erated by any point of the line. 

Call the point whose locus is desired 
P(x, y). Since the line is of fixed length, 
call the segments into which P divides it, 
x = a cosd^ y=b sin d. 

Example 2. 
a fixed line. 

Take for origin the point O where the moving point P touches the 
fixed line. If r is the radius of the circle and the angle PCD (Fig. 235) 
is 6 radians, then PD = rsind, DC = r cos 6 and OB = arc BP = rd. 



Fig. 234 

a and b (Fig. 234). Then 
Therefore the locus of P is an ellipse (§ 250). 

Find the locus of a point P on a circle which rolls along 




Fig. 235 



Now if P is denoted by the coordinates (aj, y), 

x= 0A= OB- AB= OB- PD = rd- r sine = r(d- sin d), 
y = APz=BC- DC = r-r cos d = r{l- cos d). 

Therefore 

ni\ {x = r(6— sin^), 

\y = r(l — cos d) 

are parametric equations of the curve traced by the point P. This curve 
is known as the cycloid. 



400 



MATHEMATICAL ANALYSIS [XV, § 254 



Example 3. Find the locus of a point P on a circle of radius a which 
rolls on the inside of a circle of radius 4 a. 

Take the center of the fixed circle as the origin and let the x-axis pass 
through a point M where the moving point P touches the large circle. 

Let angle MOB = d radians. Now 
we have arc PB = arc MB = 4 ad 
and arc PB = a x angle PCB. 
Therefore 

a X angle PCB = 4 ad, 
or angle PCB = 4 ^. 

But 
Z OCD + ZDCP-^Z PCB = T. 

Therefore 

|-0 + Z2)CP+4d = T, 

i.e. ZDCP=--Se. 




Fig. 236 



Now if the point P is denoted by 
{x, y) we have 

x = OE=OD-\- DE = OD-\- FP = OCcose + CP&ml--Z e\ 

= 3 a cos ^ + a cos 3 ^ = 4 a cos* ^,* 

y = EP=DG-'FC= OC sine - CPcosl--Se\ 

= 3 a sin ^ — a sin 3 ^ = 4 a sin* ^ ;♦ 
that is, 

(12) a: = 4acos3^, y = 4a8in^0. 

This curve is called the four-cusped hypocycloid. 

Example 4. P8P' is a double ordinate of an ellipse ; Q is any point 
on the curve. If QP, QP' meet the x-axis in O and 0', respectively, 
prove that CO • CO' = a^, where C is the center of the ellipse. 

Let P be (a cos ^i, ft sin di), then P' is (acos^i, — ftsin^i). Let Q 
be (a cos 6, b sin d). The equation of line PQ is 

y " sin a = —^ -(X — a cos a) 

a(cos^i — cos^) 

* Prove that cos 3 ^ = cos (2 + ^) = 4 coss — 3 cos ^, 
sin 30 = sin (2 + ^) = 3sin0— 4 8in«tf. 



XV, § 254] PARAMETRIC EQUATIONS 401 

and its x intercept {i.e. CO) is 

a(cos 6 sin d^ — sin 6 cos 6i) 
sin d — sin di 

Similarly CO' is «( - cos ^ sin ^i - sin cos gQ , 

sin ^ + sin di 

The product CO • CO' gives a"^. 

EXERCISES 

1. A line of fixed length 2 a moves with its ends always remaining on 
the coordinate axes. Find the locus of the mid-point of the line. 

2. Find the locus of the middle points of chords of an ellipse drawn 
through the positive end of the minor axis. 

3. Find the locus of a point P' on the radius CP of the cycloid (Fig. 
235) if CP' = b and b<r. 

4. The same as Ex. 3, except & > r. 

6. A circle of radius r rolls on the inside of a circle of radius a. 
Find the locus of a point P on the moving circle. 

a — r 



Ans. The hypocycloid 



X = (a — r) cos d -{- r cos 

r 

a — r 

= (a — r) sin ^ — r sin — - — < 



Ans. The epicycloid 



X = (a -\- r) cos 6 — r cos 
y = {a -h r) sin ^ — r sin 



where 6 is the same as in Example 3, § 254. 

6. A circle of radius r rolls on the outside of a circle of radius a. 
Find the locus of a point P on the moving circle. 

a + r 
a + r 

—re, 

where 6 is the same as in in Example 3, § 254. 

7. The area of the triangle inscribed in an ellipse, if ^i, ^2* ^3 are the 
eccentric angles of the vertices, is 

\ ah [sin (^2 — ^3) + sin (^3 — ^1) + sin (^1 — ^2)] 

^-2 ah sin ^2ii_^ sin h=Jl sin ^J-ZLb, 
2 2 2 

8. The coordinates of one extremity of a diameter of an ellipse are 
(a cos ^1, ft sin ^1). Show that the coordinates of one extremity of the 
conjugate diameter are (— a sin ^1, h cos ^1). 

2d 



PART IV. GENERAL ALGEBRAIC METHODS 
THE GENERAL POLYNOMIAL FUNCTION 

CHAPTER XVI 
MISCELLANEOUS ALGEBRAIC METHODS 

255. The Need of other Methods. We have hitherto con- 
sidered special functions such as a;2, sin ic, log^QX, or special 
types of functions such as mx -{- h, ax^ + bx -{-c, log„ x, a'' ; and 
we have studied their geometric and other applications. The 
study of more general types of functions, for example, the 
general polynomial of degree n, 

a^x"" + a^_iX'^~^ -h h «i^ + «o) 

of which mx + h, ax^- -f- 6a3 -f- c, ay? + bx^ -\- cx^- d are special 
types, requires more powerful methods. Some of these we 
propose to consider in the present and the succeeding chapters* 

256. Technique of Polynomials. We shall first recall the 
technique of the addition and multiplication of polynomials. 
We begin by noting that a polynomial 

can be completely represented by so called detached coeffi- 
cients, as follows : « ^ ^ « 

the place of each coefficient in this expression indicating the 
power of X to which it belongs. Thus, for example, 

2-316 
402 



XVI, § 256] ALGEBRAIC METHODS 403 

represents the polynomial 2oi:^ — Sx^ + x-\-6; and 

10 0-1 

represents the polynominal x^ — 1.* 

To add two or more polynomials we need merely add the 
coefficients of like powers of x. Thus, the sum of x"^ — l,x*-\- 
2s(r^-\-4:X^ + Sx-\-5, and 2a^ — 5x'^-\-x-\-lis given by 

10-1 
12 4 3 5 

2-5 1 1 

1 4 4 5=a;* + 4a^-f4a; + 5. 

The analogy of this process with that of adding a column of 
numbers may be noted. 

The product of two polynomials A and B is obtained by 
multiplying A by each term of B and adding the results. 
Why ? The multiplication of two polynomials by the method 
of detached coefficients is also quite analogous to the familiar 
method of multiplying two integers. Thus the product of 
2 a^ -\- Sx"^ — x —2 by aj2 + a; + 4 is given by 

23 -1 -2x114 
2 3-1-2 

2 3-1-2 

8 12-4-8 



2 5 10 9-6 -8=2a^-\-5x'-\-10a^-\-9x^-6x-S. 

The student should convince himself, by multiplying these 
polynomials in the ordinary way, that the above method is 
indeed valid. 

♦This method of representing polynomials will seem very natural, if we 
note the analogy with the familiar method of representing integers. The 
number 217 is simply a short way of writing 2 x 102 + 1 x 10 + 7, i.e. the value 
of the polynomial 2z^-{-x + 7, when x = 10. We have, therefore, been famil- 
iar with the method of detached coefficients from the time when we first 
learned to write numbers. 



404 MATHEMATICAL ANALYSIS [XYI, § 257 

257. The Division Transformation. A polynomial B is 
said to be a factor of a polynomial A, if there exists a poly- 
nomial Q such that A = BQ. A is then said to be divisible 
by B. If no such polynomial Q exists, and if the degree of B 
is less than that of A, we may always determine polynomials 
Q and E, such that 
(1) A = BQ-\-E. 

Furthermore, the remainder R can always be so determined 
that its degree is less than the degree of B. 

The process whereby a given polynominal A is expressed 
in terms of another polynomial B in the form (1), i.e. the 
process of finding Q and R, when A and B are given, is 
called the division transformation. That it is always pos- 
sible to find polynomials Q and R, if the degree of B is 
less than that of A, will be clear from the consideration of 
the following example. 

Example. Given A = 2x* + 5x^— 'Jx'^ + 12x — 5 and B = x^-10x 
+ 8 to find a polynomial Q such that ^ — J5^ is a polynomial of degree 
less than that of B. 

Since the term of highest degree in J. is 2 x* and that in B is x^, it ap- 
pears that A — 2 x^B can contain no term of degree higher than 3. In 
fact, we find A — 2 x^B = 25 x^ — 23 x^ -f 12 x — 5. Similarly, since the 
term of highest degree in ^ — 2 x^B is 25 x^, we see that the expression 
{A — 2x^B)— 25 xB can contain no term of degree higher than 2. By 
continuing this process we shall arrive at a polynomial which is of degree 
less than that of B. The work may be arranged as follows. 



^=2x* + 6 x8- 7x2-1- 12x 
J5. 2x2 = 2x4 -20x3 + 16x2 



x2-10x + 8 = 5 
2 x2 + 25 X 4- 227 



^-5.2x2 = 
J5-25x = 


25x8- 23x2-}- 12X-6 
25x8-250x2+ 200 X 


^-^(2x2 + 25x) = 
J? . 227 = 

?(2x2 4-25x + 227) = 


227x2- 188X-5 
227x2-2270x+1816 

2082x-1821 = 7 



XVI, § 258] ALGEBRAIC METHODS 405 

If the meaning of each step in the process is followed by means of the 
expressions written at the left,* it will be seen that the process has deter- 
mined a polynomial ^ = 2^^ + 23x + 227 

such that A— BQ = B, 

where B is of degree less than that of B. The nature of the process 
shows that finally such a polynomial B will always be reached. 

258. Remarks on the Division Transformation. While 
the process discussed in the last article is known as the 
division transformatioyi, it is not a process of division. Only 
if we take a further step and divide both members of the 
identity (1) (§ 257) by B, to obtain 

(2) |=0 + |, 

do we really divide A by B. The importance of this distinction 
lies in the fact that the relation (1) as derived above is valid 
without distinction for all values of the variable x involved. 
For in deriving the relation we made use only of the opera- 
tions of multiplication and subtraction. However, the relation 

(2) becomes meaningless for all values of x for which 5 = 0. 
We assumed in deriving the relation (1) that the degree 

of B was less than that of A. This is indeed necessary if Q is 
to be a proper polynomial. However, if the degree of B is 
equal to that of A, the same process will lead to a relation 

(3) A = B'q-\-R, 

where g is a constant. If the degree of B is greater than that 
of A, we may obviously write the trivial relation 

(4) A=B'()^A 

where A equals R and is by hypothesis of lower degree than B. 
If we consider a constant as a poljoiomial of degree 0, the last 

* These expressions are not of course a necessary part of the process. 
They are given here only to facilitate understanding. 



406 ' MATHEMATICAL ANALYSIS [XVI, § 258 

two cases may be included in the form (1). Our theorem then 
takes the general form 

Given any two polynomials A and B of degrees greater than 0, 
the7i polynomials Q and R can always he found such that for all 
values of the variable we have A = BQ -{- R where R is either 
zero or of degree less than that of B. 

Moreover, the transformation of A to the form BQ-\- R is 
unique, i.e. there exist just one polynomial Q and one poly- 
nomial R satisfying the conditions of the theorem. 

For, suppose there were a second pair, for example Q' and 
R'. We should then have BQ+ R = B'Q' + R', or B{Q - Q') 
= R' — R. But R' — R is either zero or of degree less than 
that of B, while B(Q— Q') is either zero or degree equal to 
or greater than that of B. Hence both are equal to zero and 
R = R', Q=Q'. 

EXERCISES 

1. Add the following polynomials by means of detached coefficients : 

(a) 2ic2 + 7x + 1, 5a;2 + 2, 3x3 + 4x-8. 
(6) 6 «2 + 5 ( + 1, 9 <2 + 8 « + 3, 6 «3 + 2 « + 1. 

(c) ay^ -\- by + c, 2 ay^ + 3by + i, 3 ay^ + 6by + 7 c. 

(d) 4 a2 -f 3 a + 2, 6 aM- 1, 4 a2 + 2 a + 3, 2 a2 + 6 a. 

2. Perform the following multiplications by means of detached coeffi- 
cients : 

(a) «3 + 2x2 + X + 3 by 2 X + 1. (c) a* + 1 by a'^ + 1. 

(b) x8 + 3x2 + 4 by x3 + X + 2. {d) y^ + 1 y + \2\)y y^ + 3y + 2. 

3. In each of the cases below transform A into the form BQ ->r B, 
where B is of degree less than B. Also write down the corresponding 
form for A/B. Detached coefficients may be used to advantage. 

(a) ^ = 6x4 4- 7 x3 - 3 x2 - 24 X - 20, B = 3 x2 + X - 6. 

(6) ^=3x4 + 2x3-32x2-66x-35, £ = x2-2x-7. 

(c) J[ = 2 x6 + 5 x8 + 13 x2 + 2 X, 5 = x2 -f 2 X + 4. 

(d) ^ = 4x7-^3x^- 19x* + 2x3-4x2-4x4-7, B = x8-x-6. 



XVI, § 259] ALGEBRAIC METHODS 407 

4. Determine m and n so that k* — moc^ -^-x"^ - nx + \ may be exactly- 
divisible by a;2 + 2 X + 1- 

6. Prove the following propositions : 

(a) If we multiply the dividend A by any constant as k{k ^ 0), we 
multiply the quotient and the remainder by k. 

(6) If we multiply the divisor by A; (A; ^t 0), we divide the quotient by 
k but leave the remainder unchanged. 

(c) If we multiply both dividend and divisor by k{k ^ 0), we multiply 
the remainder by k but leave the quotient unchanged. 

259. The Highest Common Factor of two Polynomials. 

Two polynomials A and B may or may not have a common 
factor of degree greater than 0, i.e. a polynomial F (of degree 
greater than 0) may or may not exist such that A = FQ^ 
B = FQ' where Q and Q' are also polynomials. If no such 
polynomial F of degree greater than exists, then A and B 
are said to be prime to each other. If, on the other hand, 
they have a common factor of degree greater than 0, the one of 
the highest degree is called the highest common factor (H. C. F.). 

Theorem 1. If A and B are polynomials with a common 
factor and M and N are any two polyyiomials, then any common 
factor of A and B is a factor of AM-\- BN. 

For let F be any common factor of A and B. Then we 
have A=::FQ and B=FQ'. Therefore 

AM-\- BN= F(QM+ Q'N), 

which shows that i^ is a factor of AM + BN. 

Theorem 2. If A, B, Q, R are polyyiomials such that 
A = BQ-\- B, the common factors of A and B are the same 
as the common factors of B and R. 

For, by Theorem 1, any common factor of B and i2 is a 
factor of A and hence a common factor of A and B. More- 
over, from the relation A — BQ = R and the last theorem, any 
common factor of A and B is a fac- or of B and R. 



408 MATHEMATICAL ANALYSIS [XVL § 259 

Successive applications of the division transformation there- 
fore enable us to find the H. C. F. of two polynomials A and B 
as follows : 

Using the division transformation on A and B, we may 

™*^ A = BQ + B, 

where the degree of R is less than that of B. li B = 0, B is 
the H. C. F. If 72 is a constant different from zero, then A 
and B have no H. C. F. Why ? If the degree of B is at least 
equal to 1, we may use the division transformation on B and B 
to obtain B = BQ,-hB„ 

where Bi is of degree less than B. If Bi = then B is the 
H. C. F. of ^ and B. If Bi is a constant different from zero, 
A and B are prime to each other. If B^ is of degree at least 
equal to 1, proceed as before, expressing B in the form 

B = B1Q2 -{- B2. 

This process may be continued until a remainder B^ is 
reached which is either zero, or a constant different from zero. 
If B^ = 0, then B,^_i is the H. C. F. sought. If B^ is a constant 
different from zero, then A and B are prime to each other. 

Example 1. Find the H.C. F. of 4a:8_3a;2+7a;-l and 2 a;2-3a;-|-l. 



A = ^x^ -Sx'^ + I x-1 
4a;8-6x2 + 2 x 



2x^-3x + l = B 



2x + ^ = Q 



3 x2 + 6 x-1 

3x2 -f a; + | 



Replace Bhj x — ^% — E'. 

B = 2x^-3 x + l \x-j^-g=B 

2x:2-\%x \ 2x-\i 
-nx + VA 
Therefore A and B are prime to each other. 



XVI, § 260] 



ALGEBRAIC METHODS 



409 



Example 2. Find the H. C. F. of x^ - 2 x2 -f x + 4 and 3 x^ -f 8 x2 + 
3x-2. 

The work may be arranged as follows. Since the H. C. F. of two 
polynomials is not altered in any essential way by multiplying or dividing 
either of them by any constant (^^ 0), we can avoid fractional coefficients. 



X3 
X3 


-2x2 


+ 


x-f4 

X 




-2x2 
-2x2 


+ 


2x + 4 
4-2 






21 


2x + 2 




X4-1 



3x3 


4- 


8x2 


4- 3 


X- 2 


3x3 


— 


6x^ 


+ 3 


x + 12 






14x2 


-14 


14 






X2 


-1 








X2 


+ x 








— X 


-1 








— X 


-1 







Therefore the H. C. F. is x 4- 1- 



EXERCISES 

Find the H. C. F. of each of the following pairs of expressions : 



(a) 
(&) 
(c) 
(d) 

(/) 



x3 4-2x2-13x+ 10, x3 + x2- 10x4-8. 
3x* 4- 5x2 + 2, x6 - 4 X* 4- 5 x2 - 2. 
a;8_2x2-22x + 8, x2 -6x4- 2. 



a3 + 3 a2 - 3 a - 5, aa 



a' 



4- 3 a 4- 5. 



y^ — yf^ — y + i? y'^ -\- y + i- 

&4 + 5.3 ^ 6 ^-2 _^ 5 5 ^. 5^ ^5 4. 4 53 _^ 52 _ 

1 4- x-x2- 5x3 + 4xS 1-4x84-3x4. 



6 6 4-1. 



{h) 4x4-5x3 + x + 1,3x4-4x3+1. 



(i) a^ 4- a3 + a 4- 1, a2 + a + 1. 
(j) x^>-l,x-l. 

2. Prove that, if the coefficients of two polynomials are rational (or 
real), the coefficients of the H. C. F. are rational (or real). 

3. If F is a factor of A but not of B, how does the H. C. F. of ^4 and 
FB compare with the H. C. F. of A and ^ ? In introducing and suppress- 
ing factors during the process of division, what precaution must be taken 
and why ? 

260. Functional Notation. We have already used special 
notations to represent special functions. Thus sin, cos, tan, 
log, etc. are special notations with which we are familiar. We 
shall now introduce a notation that is more general, for it is 



410 MATHEMATICAL ANALYSIS [XVI, § 260 

applicable to all kinds of functions. We shall use the symbols 
f{x)j F(x), (t>{x), — to represent various functions of x, and 
can then speak of the "/-function," " (^-function," etc., just as 
we speak of the sine function, logarithmic function, etc. 

Moreover, such a notation can be used to represent the value 
of the function in question for a given value of the variable. 
Thus, if f(x) is used to represent the function x^ -{- 2x — 1, the 
symbol /(2) denotes the value of this function when x = 2 
(just as sin (7r/2) means the value of sin x, when x = 7r/2) ; i.e. 

/(2) =22 + 2. 2-1 = 7. 

Similarly, with the meaning just given tof(x), we should have 

f(x -}-h) = {x + hy -^2ix + h)- 1. 

It should be noted that, when a certain function is called 
f(x), then, throughout any discussion where this function is 
used, f(x) always means that particular function and no other. 

EXERCISES 

1. Given/(x) = 3x2 + 2a;- 4, fiiid/(2),/(^),/(0),/(a; + 1/x). 

2. Given 0(x) = x/(x - 1), find 0(2), (f>(x + h), 0(1 - x), 0(10). 

3. Given F{x) = e* + e-^ find F(0), F(l). 

4. Given/(x) = (x - l)/(a; + 1), prove that 

6. If 0(x + y) = (t>{x) + 0(j/), show that 0(3 x) = 3 0(a;). 



6. Given f(x) = 2 x Vl — x^, prove that 

/(sinx) =/(cosx) =sin2x. 

1 



X 

Given /(x) = -, find the value of /Ml+^V 



X 



8. Given ^(x) = e* + e"*, prove that 

e{x^y)d{x-y) =^(2x) -}-^(2y). 

9. Express tlie fact that the volume of a sphere is a function of its radius. 



XVI, § 262] ALGEBRAIC METHODS 411 

10. Given F {x) = (x- \/x){x'^ - \/x'^){xJ^ - \/x?), prove that 

F{z) = -F{\/z), 

11. Given that/(w) = n !,* prove that (n + 1) f{n) =f(n + 1). 

12. Given that/(.7;)=a;2-f-2, find/[/(x)]. 

13. Given /(x) = sinx, find/(7r/2),/(7r/3), /(tt). 

14. Given /(x) = sin x and <f)(x) = cos x, prove that 

(«) lf(x)T + [0(^)]-^ - 1. (d) fix) = «/,(7r/2 - X). 

(b) f(x) - 0(a-) = tan X. (e) /(- x) = -/(x). 

(c) /(.r + 2/) =f(x) cf>(y) +f(y) <p{x). (/) 0(x) = 0( - x). 

16. If /(x) = loga X, show that 

(a) / (a;) + / (y) =f{xy). (c) /(xA) + f(y/x) = 0. 

(6) /(x") = n/(x). 

16. What functions may/(x) represent when 

(a) fix + y) =f(ix)f(y). (c) /(x") = n/(x). 

(6) /(x+ 2/) =/(x) +/(2/). (d) /(^^ =/(2/) -/(x). 

261. The Remainder Theorem. If a polynomial f(x) is 
divided by x— a, the remainder isf(a). 

If f(x) is the dividend, a; — a is the divisor, Q(x) the 
quotient, and R the remainder, then 

(5) f{x)=(x-a)Q(x)-^E 

where B, since it is of lower degree than x — a, does not in- 
volve X at all, i. e. E has the same value for all values of x. 

Since the values of the two members of this identity are 
equal to each other for all values of x, we have for the par- 
ticular value X = a. ^/ s r> 

J {a) = M. 

262. Factor Theorem. If f(x) is a polynomial and a is a 
number such that f (a) = 0, then x — a is a factor off(x). 

The proof of this theorem is left as an exercise. 

*n! = l-2.3.4 •... • w, that is, 2! =2, 3! =6, 4! =24, .... 



412 MATHEMATICAL ANALYSIS [XVI, § 262 

EXERCISES 

By use of the remainder theorem find the remainder when 

1. x^ — 2 x'^ + 3 is divided by a; — 2. 

2. a:i3 _ 45 a:i2 + 26 x^ + 12 is divided by a; - 1. 

3. a:i2 + 1 is divided by jc + l ; by a; — 1. 

4. Show that — 2 is a root of the equation 2 x^ + 3 a;^ — 4a; — 4 = 0. 

5. Show that a;'* + «" is divisible by x + a if w is odd. 

6. Show that x" + a" is not divisible by x + a if n is even. 

7. By means of the remainder theorem find k so that x^-\- kx^-\-2x +^ 
may be exactly divisible by a; — 2. 

8. Find the polynomial in x of the second degree which vanishes when 
x= I and when a; = 2, and which assumes the value 10 when x = 3. 

263. Synthetic Division. We shall proceed to explain a 
simple method of performing the division transformation 
when the divisor has the form x — k, i.e. when the divisor is a 
binomial of the first degree in which the leading coefficient is 1. 

Let the given polynomial be 

a„aj" + «„-iaJ"~^ + a^-2X"'-^ + ••• + OjX + a^ 
and the divisor x — k. 

The ordinary process of long division leads to 

X — k 



a,.a;^— a„ A;.r"-i ^«^"~^ + (^«n + ^n-i)^""^ 

(ka, + ft^ -Qa;"-^ - k(ka^ + a.,-i)x^-^ 

It is now not difficult to see that 

(a) the first coefficient in the quotient is a„, i.e. the coeffi- 
cient of the leading term in the dividend ; 

(b) the second coefficient in the quotient is obtained by 
multiplying the first coefficient of the quotient by k and adding 
to it the second coefficient of the dividend : 



XVI, § 263] ALGEBRAIC METHODS 413 

(c) the third coefficient of the quotient is obtained by mul- 
tiplying the second coefficient of the quotient by k and adding 
to it the second coefficient of the dividend. 

We may then arrange the work as follows : 

a„ a„_i a„_2 «„-3 - 1^ 

or„ ka^ -f a„_i k{ka„ + a„_i) + a„_2 - 

Here the first line contains the coefficients of the dividend 
in order and the third line gives the coefficients of the quotient 
and the remainder in order. Every number in the third line, 
after the first, is obtained by multiplying the preceding num- 
ber by k and adding to it the next number in the first line. 

Example 1. By synthetic division, find the quotient and the re- 
mainder when X*— 2ic3-f-x2 + 3a;— 2is divided by cc + 2. 

Solution : 1-21 3—21 — 2 

- 2 8 - 18 30 
1 _ 4 9 - 15 28 
Hence the quotient is x^ — 4 ic^ + 9 x — 15 and the remainder is 28. 

Example 2. If/(x) = 2x* +3x3+7x2-}- 14x + 20, find/(-3). 

2 3 7 14 201-3 

-6 9 -48 102 



2 -3 16 -34 |122=/(-3). Ans. 

EXERCISES 

In the following exercises use synthetic division : 

1. Find the remainder when x^ + 3 x^ — 6 x + 2 is divided by x — 2. 

2. Find the remainder and the quotient when x* — 3x'^ + 2x + 3 is 
divided by x + 3. 

3. Show that 3 is a root of the equation x^ - 4 x^ - 17 x + 60 = 0. 

4. Find k so that 3 is a root of the equation x'* - 3 x^ + A:x + 1 = 0. 

5. Is 5 a root of the equation x^ — x^ + 7 = ? 



414 MATHEMATICAL ANALYSIS [XVI, § 264 

264. Properties of Integers. We have already noticed 
(ftn.j p. 403) that the familiar method of writing an integer 
greater than 9 represents it as the value of a polynomial. 
Integers and polynomials, therefore, have many properties in 
common. We may, for example, gain an insight into the reason 
for the rules of arithmetic used in adding a column of figures 
or in finding the product of two integers by comparing these 
rules with the technique of adding and multiplying polynomials 
discussed in § 256 *. We shall proceed to discuss some of the 
properties of integers relating to divisibility, etc., which are 
valuable in our everyday use of numbers. 

265. Prime and Composite Numbers. An integer greater 

than 1 that is divisible by no integer except itself and 1 is called a prime 
number or simply a prime. Thus 2, 3, 5, 7, 13 are prime numbers. Any 
integer (> 1) not a prime is called a composite number. Any composite 
number is the product of two or more primes, thus 6 = 2-3, 100 = 2 • 2 • 
5 . 5 = 22 . 5^. Any composite number 7i may be resolved into its prime 
factors in one and only one way. When resolved it has the form 
n = j?i*ij92"* •••Pfc"*- When a number has been resolved into its prime 
factors any question regarding its divisibility is readily answered by the 
following theorem. 

A number a is divisible by a number b if and only if every prime factor 
of b occurs in a to at least as high a power as it occurs in b. This 
theorem follows readily from exercises 15 and 17 below. The proof is 
left to the student. As an illustration, if a = 2 • 3^ • 17^ • 37 and 6 = 
2 • 32 . 17 we recognize at once that a is divisible by b and that the quo- 
tient is 3 • 17 • 37. If, on the other hand, b were 2^ . 3^ • 17 then a would 
not be divisible by &. 

The common factors of two integers are also readily found if the num- 
bers have been resolved into their prime factors. Why ? Two integers 
which have no common factor (> 1) are said to be prime to each other. 

The notion of prime numbers and the investigation of their properties 
is very ancient and to this day some of the most difficult problems of 
advanced mathematics relate to this field. Some of the properties are 
quite elementary, however, and have been listed below in exercises. 

* Carrying this comparison out in detail forms a valuable exercise. The 
familiar process of "carrying" a digit from one column to the next is about 
the only thing that requires special investigation. 



XVI, § 265] ALGEBRAIC METHODS 415 

EXERCISES 

1. Prove that if a and h are both divisible by w, then a-\-h and a—b 
are divisible by n and a • 6 is divisible by n^. Is a similar theorem true 
of polynomials ? 

2. Prove that a product of any number of integers is divisible by n if 
one of the integers is divisible by n. Is a similar theorem true of poly- 
nomials ? 

3. If c = a & is divisible by «, must either a or 6 be divisible by n ? 

4. Prove that if a number a leaves a remainder r when divided by &, 
then the number obtained by adding to a any multiple of h will leave the 
same remainder. 

5. Prove that if the last digit on the right of an integer is even, the 
integer is divisible by 2. 

6. Prove that if the number formed by the last two digits of an 
integer is divisible by 4, then the number is divisible by 4. 

7. Prove that if the number formed by the last three digits of an 
integer is divisible by 8, then the integer is divisible by 8. 

8. Prove that if the last digit of an integer is or 5 then the integer 
is divisible by 5. 

9. Prove that if the sum of the digits of an integer is divisible by 3 
(or 9) then the integer is divisible by 3 (or 9). 

10. Prove that if the sum of the first, third, fifth, etc. digits of an 
integer is equal to the sum of the second, fourth, etc., the number is 
divisible by 11. 

11. If the sum of the digits of an even number is divisible by 3, the 
number itself is divisible by 6. 

12. Determine without performing the division whether the following 
numbers are divisible by 2, 3, 4, 5, 6, 8, 9, 11. 

(a) 2562. (c) 123453. (e) 127898712. {g) 111111111111. 

(6) 12345. {d) 2673. (/) 7325 x 8931. Qi) 11111111112. 

13. How would you recognize that a number is divisible by 45 ? 

14. Prove that if the product « • & is divisible by a prime number p, 
either a is divisible by p or 6 is divisible by p. Is a similar theorem true 
for polynomials ? 

15. Prove that if a number c is a factor of ah and is prime to a, it 
must be a factor of h. Is a similar theorem true for polynomials ? 

16. Prove that the quotients of two numbers by their H. C. F. are two 
numbers prime to each other. Is a similar theorem true for polynomials ? 



416 MATHEMATICAL ANALYSIS XVI, § 265 

17. Show that if a number is divisible by each of two numbers which 
are prime to each other, it is divisible by their product. Is a similar theo- 
rem true for polynomials ? 

18. Show that the product of two numbers is equal to the product of 
their H. C. F. by their L. C. M. Is a similar theorem true of polynomials ? 

19. Prove that the number of primes is unlimited. 

[Hint. Suppose that the theorem were not true and that p were the 

greatest prime. Let pi, /)2, P3, •••, Pn-i be the set of all primes less than p 

and consider the number 

PiP'zPz ■■' Pn-iP-^ 1- 

This number is not divisible by any of the primes pi, p2, •••, p. The rest 
of the proof should be obvious. This proof was first given by Euclid.] 

20. By trying successive primes 2, 3, 5, 7, •••, determine whetlier or not 
1009 and 1007 are primes. In this case we may stop with the prime 31. 
Wiiy ? Ans. 1009 is prime. 

• 21. Resolve into prime factors the numbers 604800 and 12250. 

22. Is the number 2^31253 a perfect square ? Is it a perfect cube ? 

23. Show that the relation ah — cd = 1, where a, &, c, rf, represent in- 
tegers, is not possible unless a and c are prime to each other. 

24. Two consecutive integers are always prime to each other. Is this 
true also of any two numbers differing by 7 ? 

25. What is the smallest cube of masonry that can be constructed of 
bricks 8x3x2 inches ? It is assumed that the bricks are placed so that 
any two equal sides are parallel. 

266. Partial Fractions, in certain problems it is sometimes found 
necessary to express a fraction in which the numerator and denominator 
are polynomials in one variable as the sum of several fractions each of 
which has a linear or at most a quadratic function in the denominator. 
In what follows it will always be assumed that the given fraction is a 
proper fraction^ i.e. a fraction in which the degree of the numerator is less 
than that of the denominator. Any fraction which is not proper can be 
written as the sum of a polynomial and a proper fraction. Therefore our 
problem may be stated as follows : To express a proper fraction as the 
sum of several proper fractions. 

The method of approach is to assume that the fraction can be expressed 
in the desired form and then seek to determine the numerators which in 
the assumed form are left undetermined. Four distinct cases arise. 



XVI, § 266] ALGEBRAIC METHODS 417 

Case I. When the denominator can be resolved into factors of the first 
degree all of which are real and distinct. 
Example 1. Resolve into partial fractions 
9x^ -a;-2 
x^ — X 
The sum of the three fractions 

X X —\ X -\-l 

will give a fraction whose denominator is x^ — x. We therefore try tc 
determine A^ B, C so that 

(Q) 9x^-x-2 _A B C 

X^ — X X X — 1 x -\- 1 

Clearing of fractions we have 
(7) 9x-2-x-2 = A{x^ - 1) + B{x^ ■{■ x) -{- C {x'^ - x) . 

Since (7) is to be true for all values of x, we seek values of A, B, O, such 
that the coefficients of like powers of x will be equal, i.e. such that 

A + B-\- C^9, B-C = -l, -A =-2. 

Solving these equations, we find A = 2, B = S, (7 = 4. Hence 
9a;2-a;-2_2 3 , 4 



X^ — X x X—1 X -\- I 

Case II. When the denominator can be resolved into real linear fac- 
tors some of which are repeated. 

Example 2. Resolve into partial fractions 
2 a;2 - X + 2 
x(x-iy 
The sum of the fractions 

A^ B , C 



X x-\ {X- 1)2 

will give a fraction whose denominator is x{x — l)^. Therefore we shall 
try to determine A, B, C so that 

2x^-x + 2^A^ B , C 



X{X - 1)2 X X-l {X- 1)2 

Clearing of fractions, we have 

2x2-3-4-2 = A{x- 1)2 J^ Bx{x- 1)+ Cx. 
Equating the coefficients of like powers of x, we have 

A-^B = 2, -2A-B-\-C = -\, A = 2. 
2e 



418 MATHEMATICAL ANALYSIS [XVI, § 266 

Solving for A, B, O, we find ^ = 2, B = 0, C = 3. Hence 

2x2-a;+2_2 3 



x(a'-l)2 X, (a;-l)2 

The assumptions to be made under Cases I and II are contained in the 
following rules. 

(1) For every unrepealed factor x— a of the denominator^ assume a 
fraction of the form A/{x — a). 

(2) For every repeated factor {x — a)* of the denominator^ assume a 
sum of fractions of the form 

A\ ^ A2 _^ ,,, I Ak 
X — a (x— a)2 (x — ay 

Case III. W?ien the denominator contains quadratic factors which 
are not repeated. 

Example 3. Resolve into partial fractions 
5 x2 + 4 X + 3 

(X+1)(X2 + 1)' 

Let 5 x2 + 4 X + 3 ^ A Bx + G 

(x+ l)(x2+ 1) x + 1 x-^+ 1 ' 

Clearing of fractions, we have 

5 a;-2 + 4 X + 3 = A(x^ +1) -{-(Bx+ C)(x + 1) 

= Ax^ + A -{- Bx'^ -^ Bx + Cx + a 
Equating coefficients, 

A-^B=o, B+C = 4,A+C=S. 

Therefore A = 2, B = S, 0=1. Hence 

5x^ + 4x+3 ^ 2 . 3x4-1 
(x+l)(x2 + l) x+1 X24-1' 

Case IV. When the denominator contains quadratic factors which 
are repeated. 

Example. Resolve into partial fractions 

3 x^ + x'^ + 8 x'-! + X + 2 

X(X2 + 1)2 

Let 3 x^ + x^ + 8 x2 + a; + 2 ^A Bx + C Dx + E 
x(x2 +1)2 X x2 + 1 (x2 -f 1)^' 

Then, 
3 x* + x8 + 8 x2 + X + 2 = .4(x2 + l)a + (^x + C)x(x2 + 1) + {Dx + E)x 
= ^x* + 2 ^x2 + A + Bx^ + Cx3 + ^x2 + Cx + Z)x2 + Ex. 



XVI, § 266] ALGEBRAIC METHODS 419 

Equating coefi&cients we have 

^ + JB = 3, C=l, 2^ + 5 + i) = 8, C+E = l, A = 2. 
Hence, A = 2, B = 1, 0=1, D = 3, E = 0. 

Therefore, ^^^ ^ x^ + Sx'^ + x + 2 ^2 x±l_ Sx 

x(x^+iy^ X a:2 -f 1 (a;2 + 1)2 

The assumptions to be made in Cases III and IV are contained in the 
following rules. 

3. Corresponding to every unrepealed factor of the form ax'^ + hx -\- c, 
assume the partial fraction Ax A- B 

ax?- + 5x + c 

4. Corresponding to evet^ factor (ax^ -\- bx + c)*, assume the sum 

Aix + Bi J A2X + B2 _|_ ^^^ AkX 4- Bk 

ax^ -^ bx -\- c (aa;2 -f 6x + c)2 (q[x2 + 6a: + c)*' 

EXERCISES 

Resolve into partial fractions each of the following fractions. 
J 4x+ 1 jj 3 a:2 - 5 X + 4 



(a;_l)(x+l)(x + 3) 
3a;-l 
x2-4 ' 
2x + l 



x2(x — 4) 
x^ + 1 

X(X- 1)8* 

1 

x%x + 1) * 
x2 + 2 a; + 1 

x^ + x 
2 x2 - 1 
3x3 + 3 x' 

2x + l 

X3 + X2 + X * 
1 



12. 
13. 
14. 
16. 
16. 
17. 
18. 
19 



X(X2 + 1)2 

3 

X8-1' '" X2(X2+1)2 



10. -^. 20 



(X - 1)3 
X2 


(X2-1)2' 
X* 


(x2-l)(x + 2) 
x-2 


(X + 1)X2 

2 x2 - X + 3 


x(x2-l)(2x-3) 
1 +x^ 


(2 - x2) (2 + x2) ' 

X* + X2 


X* + X2 + 1 

3-2x2 


(2 - 3 X + x2)2 
5 x3 4- 2 X + 1 


(X^+1)(X-1)2° 

2x+l 



CHAPTER XVII 

PERMUTATIONS, COMBINATIONS, AND PROBABILITY 
THE BINOMIAL THEOREM 

267. Definitions. Suppose that a group of n objects is 
given. Any set of r (r ^ n) of these objects, considered with- 
out regard to order, is called a combination of the n objects 
taJce7i r at a time. We often denote the objects in question, 
which may be of any kind, by letters, as a, 6, c, •••, k. The 
number of combinations of these n letters taken r at a time is 
denoted by the symbol „C^. For example, the combinations 
two at a time of the four letters a, b, c, d are, 

ab, ac, ad, be, bd, cd. 

Since there are 6 of these combinations in all, we have 4C2 = 6. 

On the other hand, any arrangement of r of these n objects 
in a definite order in a row is called a permutation of the n 
objects taken r at a time. The symbol „P^ is used to denote 
the number of such permutations. 

For example, the permutations of the four letters a, b, c, d 
taken two at a time are 

ab ac ad be bd cd ba ea da cb db dc. 

Since there are 12 of these arrangements in all we have 
4P2 = 12. We have assumed in these examples that the 
objects are all different, and that the repetition of a letter 
within a permutation is not allowed. 

268. Fundamental Principle. If a certain thing can be 
done in m different ways and if, when it has been done, a cer- 
tain other thing can be done in p different ways, then both 

420 



XVII, §269] PERMUTATIONS AND COMBINATIONS 421 

things can be done in the order stated in m x p different 
ways. For, corresponding to the first way of doing the first 
thing, there are p different ways of doing the second thing ; 
corresponding to the second way of doing the first thing there 
are p different ways of doing the second thing ; and so on for 
each of the m different ways of doing the first thing. There- 
fore there are m x p different ways of doing both things in 
the order stated. This fundamental principle may at once be 
extended to the following form. 

If one thing can be done in m ivays, and if, when it has been 
done, a second can be done in p loays, and if when that has been 
done, a third can be done in q ways, and so forth, then the number 
of ways ill which they can all be done, taking them in the order 
stated, is m X p xq •••. 

Example 1. There are five trails leading to the top of Mt. Moosilauke, 
N. H. In how many ways may I go to the top, and return by a different 
trail ? 

There are five ways I may go to the top and for each of these there are 
four ways I may descend. Therefore, the total number of ways in which 
I may make the round trip is 5 x 4 or 20. 

Example 2. How many even numbers of two unlike digits can be 
formed with the digits 1, 2, 3, 4, 5, 6, 7, 8, 9 ? 

The digit in the units' place can be chosen in any one of 4 ways and the 
one in the tens' place can then be chosen in 8 ways. Tlierefore, 4 x 8 or 32 
even numbers with two unlike digits can be formed from the given digits. 

269. The Number of Permutations of n Different Things 
Taken r at a Time. The problem of finding the number of 
permutations of n different things taken r at a time can be 
stated as follows : 

Find the number of ways in which we can fill r places when 
we have n different things at our disposal. 

The first place can be filled in n ways, for we may take any 
one of the n things at our disposal. The second place can 



422 MATHEMATICAL ANALYSIS [XVII, § 269 

then be filled in ?i — 1 ways, and hence the first and second 
places together can be filled in n {n — 1) ways. Why ? When 
the first two places are filled, the third can be filled in w — 2 
ways. Reasoning as before, we have that the first three places 
can be filled in n(n — l)(n — 2) ways. Proceeding thus, we see 
that the number of ways in which r places can be filled is 

n (ii — l)(?i — 2) ..• to r factors, 

and the rth factor is n — (r — l) or n — r + 1. Therefore the num- 
ber of permutations of n different things taken r at a time is 

(1) ^p^ = „(„_i)(n-2)...(n-r + l). 

Corollary. If r = n, we have 

(2) ^Pn = n (n - 1) (n - 2) ... 3 . 2 . 1 = n ! * 

Example. Three students enter an oflBce in which there are five 
vacant chairs. In how many ways can they be seated ? 
Here n = 6, r = 3. Hence sPs = 6 • 4 • 3 = 60 ways. 

270. The Permutations of n Things not all Different. The 

number N of permutations of n things taken all at a time, of 
which p are alike, q others are alike, r others alike, and so on, is 

(3) N= ~ ■' 

Suppose the n things are letters and that p of them are a, 
q of them h, r of them c, and so on. 

Now, if in any of the N permutations we replace the p a's 
by p new letters, different from each other and also from the 
remaining n — p letters, then by permuting these p letters 
among themselves without changing the position of any of the 
other letters we can form p ! new permutations. Therefore if 
this were done in each of the N permutations, we should 

* The product of all the integers from 1 to n is called factorial n, and is de- 
noted by the symbol n ! or [n. Thus 3 ! = 1 • 2 • 3 = 6. A table of the values 
of n ! up to n = 10 will be found at the end of the book. 



XVII, § 270] PERMUTATIONS AND COMBINATIONS 423 

obtain N-pl new permutations. In the same manner, if we 
replace the q 6's by q new letters differing from each other and 
the remaining n — q letters, the r c's by r new letters differing 
from each other and from the remaining n — r letters, and so 
on, we then obtain N -plqlrl "> new permutations. But the 
things are now all different and may be permuted in n I 
ways. Therefore J^ - p\ - q I - r \ "■ = nl, or 



p Iqlrl '" 

Example. How many different permutations of the letters of the 
word Mississippi can be formed taking the letters all together ? 

We have 11 letters of which 4 are s, 2 are p, 4 are i. Therefore the 
number of permutations is 11 !/(4 ! 4 ! 2 !) = 34650. 

EXERCISES 

1. If there are six letter boxes, in how many ways can two letters be 
posted if they are not both posted in the same box ? Ans. 30. 

2. If there are six letter boxes, in how many ways can two letters be 
posted ? Ans. 36. 

3. Two dice are thrown on a table. In how many ways can they 
fall? Ans. 36. 

4. Two coins are tossed on a table. In how many ways can they fall ? 

5. In how many ways can five coins fall on a table ? 

6. How many different permutations can be formed by taking five 
of the letters of the word compare ? 

7. Find the number of permutations that can be made from all the 
letters of the word (a) assassination; (b) institutions; (c) examination. 

8. Given the digits 1, 2, 3, 4, 5, 6, 7, 8, 9. Find 

(a) How many odd numbers of two digits each can be formed, repeti- 
tion of digits being allowed. 

(b) The same as (a), except that repetition of digits is not allowed. 

9. How many even numbers less than 1000 can be formed with the 
digits 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, repetition of digits not being allowed ? 

10. In how many ways can a hand of ten cards be played one card 
at a time ? 

11. In how many ways can 3 different algebras and 4 different geom- 
etries be arranged on a shelf so that the algebras are together? 



424 MATHEMATICAL ANALYSIS [XVII, § 271 

271. The Number of Combinations of n Different Things 
Taken r at a Time. TJie number of combinations of n different 
things taken r at a time is 

(4) „C, = "("-^)("-^)-("-'- + ^). 

r ! 

For, each combination consists of a group of r different 
things which can be arranged among themselves in r ! ways. 
Therefore ^C^ • r I is equal to the number of permutations of n 
different things taken r at a time ; that is, „C^ • r ! = „P,, or 

^ _ n(n — l)(n — 2) >» (n — r + 1) 
-^~ r! 

Corollary 1. The value of „(7, may be written in the form 

rl{n — r)l 

This follows immediately from (4) if we multiply numer- 
ator and denominator by (n — r) !, since 

n(n- l)(n - 2) ... (w - r + 1) • (n - r) ! = n!. 

Corollary 2. The number of combinations of n different 
things taken r at a time is equal to the number of combina- 
tions of n different things taken (n — r) at a time. 



= „a. 



(n — r)l (??, — [n — r]) ! (n —r) I r\ 

The total number of ways in which a selection of some or all 
can be made from n different things is 2" — 1. For each thing 
may be disposed of in two ways, i.e. it may be taken or it may 
be left. Since there are n things, they may all be disposed of in 
2" ways. But among these 2" ways is included the case in 
which all are rejected. Therefore the number of ways of 
making the selection is 2" — 1. 



XVII, § 271J PERMUTATIONS AND COMBINATIONS 425 

Example 1. In how many ways can a committee of 9 be chosen from 
12 people ? 

The required number is 

^ _ ^ _ 12 . 11 . 10 _ „„^ 
12t^9 — 12^3 — — q — r — — — ^IM. 

Example 2. From 6 men and 5 women, how many committees of 8 
each can be formed when the committee contains (1) exactly 3 women ? 
(2) at least two women ? 

(1) The men may be chosen in eCs ways, the women in gCs ways. 
The number of ways in wliich both groups may be chosen together is 
eCs-eCs, or60. 

(2) Since each committee is to contain at least three women, it can be 

made up as follows : , . r ■, ^ 

(a) 5 men and 3 women. 

(b) 4 men and 4 women. 

(c) 3 men and 5 women. 

Therefore the number of possible committees is « 

6^6 X 6<73 + 6^4 X 6^4 + 6C3 X 5^5 = 165. 

EXERCISES 

1. Find 10 Cg; 11 10; 100C99. 

2. How many different committees of 6 men can be chosen from a 
group of 20 men ? 

3. There are 20 points in a plane, of which no three are in a straight 
line. How many triangles may be formed each of which has three of 
these points for its vertices ? 

4. How many planes may be determined by 25 points, no four of which 
are coplanar, if each of the planes is to contain three points ? 

5. How many different committees, each consisting of 5 republicans 
and 4 democrats, can be formed from 10 republicans and 8 democrats ? 

6. From 20 men how many groups of 11 men each can be picked ? In 
how many of these groups will any given one of the 11 men be ? 

7. Out of 6 different consonants and 4 different vowels, how many 
linear arrangements of letters each containing 4 consonants and 3 vowels 
can be formed? Ans. ^Ga x 403 x 7 !. 

8. From ten books, in how many ways can a selection of six be made, 

(1) when a specified book is always included ? 

(2) when a specified book is always excluded ? 



426 MATHEMATICAL ANALYSIS [XVII, § 272 

272. Probability. If an event can happen in h ways and fail 
in / ways, and if each of these f+h ways is equally likely, the 
(mathematical) probability * of the event happening is 

h 
h-\-f 

and the probability of its failing is f/{h +/). An equivalent 
way of stating that h/(Ji -\-f) is the probability of an event 
happening is to say that the odds are h to / in favor of the 
event or / to 7i against the event. 

The probability of an event happening plus the probability of 
its failing is always equal to unity. 

Example 1. Suppose from a bag containing 3 red balls and 5 black 
ones, a ball is drawn at random, then the probability of its being red is f 
and of its being black f . The chance that the ball is either red or black 
is I + f = 1, or certainty. 

Example 2. From a bag containing 3 red balls and 6 black ones, two 
balls are drawn. Find the probability that (1) both are red, (2) both are 
black, (3) one is red and one is black. 

Two balls can be drawn in g C^ or 28 ways. Two red balls can be drawn 
in 3C2 or 3 ways. Therefore the probability of drawing two red balls is 
3/28. 

Two black balls can be drawn in 5 (7-2 or 10 ways. Therefore the prob- 
ability of drawing two black balls is 10/28. 

The number of ways of drawing one red ball and one black one is 
zCi X 6^1, or 15. Therefore the probability of drawing a red and a black 
ball is 15/28. 

Example 3. Find the probability of throwing six with two dice. The 
total number of ways in which two dice can fall is 6 x 6 or 36. A throw 
of 6 can be made as follows: 1, 6; 5,1; 4,2; 2,4; 3,3; i.e. in 
5 ways. Therefore the probability is 5/36. 

* The reason for the definition of mathematical probability may be made 
clear from the following considerations. Suppose a coin were tossed vi times 
and fell heads h times and tails/ times. If n is a finite number, h and/ will 
in general not be equal. But as n is increased, h/{h+f) and f/(h-\-f) will 
approach nearer and nearer to 1/2, and thus we take 1/2 to be the probability 
of the coin falling heads 



XVII, § 272] PROBABILITY 427 

EXERCISES 

1. In a single throw with one die, find the probability of throwing an ace. 

2. In a single throw with two dice, find the probability of throwing a 
total of five ; six ; seven ; eight. 

3. In a single throw with two dice, find the probability of throwing at 
least five ; six ; seven ; eight. 

4. A bag contains 5 red balls, 6 green balls, 10 blue balls. Find the 
probability that, if 6 balls are drawn, they are (a) 2 red, 2 green, 2 blue; 

(b) 3 green, 3 blue; (c) 5 red, 1 green ; (d) 6 blue. 

5. Four coins are tossed. Find the probability that they fall two heads 
and two tails. Ans. 3/8. 

6. In a throw with two dice, which sura is more likely to be thrown, 
6 or 9 ? 

7. Find the probability of throwing doublets in a throw with two dice. 

8. Five cards are drawn from a pack of 52. Find the probability that 
(«) there is one pair. [Two like denominations make a pair, for ex- 
ample, two aces.] 

(6) Find the probability that there are three of a kind ; (c) two pairs ; 
(d) three of a kind and a pair ; (e) four of a kind ; (/) five cards of one 
suit. 

9. Four cards are drawn from a pack of 52. Find the probability 
that they are one of each suit. 

10. Seven boys stand in line. Find the probability that (a) a partic- 
ular boy will stand at an end; (6) two particular boys will be together ; 

(c) a particular boy will be in the middle. 

11. A and B each throw two dice. If A throws 8, find the probability 
that B will throw a higher number. 

12. Find the probability of throwing two 6's and one 5 in a single 
throw with three dice. * 

13. In tossing three coins find the probability that at least two will be 
heads. 

14. If the probability that I shall win a certain event is |, what are the 
odds in my favor ? 

15. Find the probability of throwing an ace with a single throw of two 
dice. Ans. 11/36. 

16. Which is more likely to happen, a throw of 4 with one die or a 
throw of 8 with two dice ? 



428 MATHEMATICAL ANALYSIS [XVII, § 273 

273. The Binomial Theorem for Positive Integral Ex- 
ponents. Consider the product 

(x -\-a)(x 4- a) ••• (x + a) [to n factors] 

where n is any positive integer. One term' of the product is 
oj" ; it is obtained by taking the letter x from each parenthesis. 
There will be n terms x^'^a, for the letter a may be chosen from 
any of the n parentheses which can be done in „0i = n ways. 
There will be „(72 terms x^'^a"^, for the a's may be chosen from 
two of the n parentheses and the x from the remaining n — 2 
parentheses. In general, there will be nO^ terms a;""'" a**, for 
the a's may be chosen from any r of the n parentheses, and the 
x's from the remaining n — r parentheses. Therefore 

(6) (x + a)^ =zx^ + nCiX'»-ia + nCax'*-^ a^ + ... 

-\-„CrX'^-rar + ... + a". 

This formula for expanding (x -\- ay is known as the binomial 
theorem. Since „(7^ = ^(7„_„ it follows that the coefficients of 
any two terms equidistant from the beginning and the end are 
equal. If we write — a in place of a we have 

(x - ay = x-~' + nCiX--' (- a) + n^aaj'*'' (- ay + ... -f (- a)% 

or 

(aj-a)« = aj"— „Oia.-"-ia+ ^OgX^-^a^— ^CaOJ^-Vi^ -j 1- ( _ 1)" a\ 

Example 1. Expand (2x — yy. 

= 32 x5 - 80 x^y + 80 xV _ 40 x'Y + 10 a-y^ _ yS^ 

Example 2. Find the sixth term of (2 as — 3 yy. 
The sixth term is gCs (2xY ( - 3^)5, or - 108,864 x^. 

Example 3. Find what term contains x" in the expansion of ( x^ — J . 

Call it the «"i term. ThenioC«_i(x2)"-<^-iy~^ is the term. In this term 

we want the exponent of x to be 11. Therefore 22 — 2«— «-f-l = ll, or 
t = 4. The coefficient of this term is - 10C3 = — 120. 



XVII, § 273] BINOMIAL THEOREM 429 

EXERCISES 

Expand the following by the binomial theorem : 

1. (x-l)5. 3. {2x-yy. 5. (x--y, 

/ l\io ^ ^' 

2. (2x + y)6. 4- [^--) • 6. {z - xy)K 

7. (0.9)6. [-Hint. 0.9 = 1-0.1.] 8. (0.99)3. 

Write down and simplify : 

9.. The 8th term of (x - iy\ 12. The 6th term of (2 a; + 3 y)i2. 

10. The 5th term of (2 x - y)io. 13- The middle term of (1 - xy\ 

U. The middle term of (2 x - y)i*. 

15. The middle terms of {z - \/z) i^. 



11. The 7th term 
Find the coefficient of 



\b ix) 



16. x^ in the expansion of (x^ — -j , 

17. a;i8 in the expansion of [x^-l--] . 



1 \ 



15 



18. x^9 in the expansion of ( x* +- i . 

19. x-^'' in the expansion of ( x* — ^ ] . 

V x^J 

20. By considering the expansion of (1 + 1)», prove that 

21. Prove 1 _ „Ci + „(72 - „C3 + •.- + (- 1)" „Cn = 0. 

MISCELLANEOUS EXERCISES 

1. In how many ways can 10 boys stand in a row ? 

2. In how many ways can ten boys stand in a row when 

(a) A given boy is always at a given end ? 
(6) A given boy is always at an end ? 

(c) Two given boys are always together ? 

(d) Two given boys are never together ? 

3. How many numbers of three digits each can be formed from the 
digits 1, 2, 3, 4, 5, 6, 7, when 

(a) A repetition of digits is allowed ? 
(6) A repetition of digits is not allowed ? 



430 MATHEMATICAL ANALYSIS [XVII, § 273 

4. How many numbers of three digits each can be formed with the 
digits 2, 3, 5, 6, 7, 9, when 

(a) The numbers are less than 500 and a repetition of digits is 
allowed ? 

(&) The numbers are greater than 500 and a repetition of digits is 
not allowed ? 

5. In how many ways can a consonant and a vowel be chosen from 
the letters of the word vowels ? 

6. Find n when, 

(a)„(72 = 45; (6) „P3 .= 210 ; (c)„(72 = „C3. * 

7. Show that the number of ways in which n things can be arranged 
around a circle is (n — 3) ! . 

8. In how many ways can 6 people sit around a round table ? 

9. How many signals can be made by hoisting 7 flags all at a time one 
above the other, if 2 are blue, 3 are white, and the rest are green ? 

10. How many different numbers of seven digits each can be formed 
with the digits 1, 2, 3, 4, 3, 2, 1, the second, fourth, and sixth digits being 
even? 

11. How many handshakes may be exchanged among a party of 10 
students if no two students shake hands with each other more than once ? 

12. A lodge has 50 members of whom 6 are physicians. In how many 
ways can a committee of 10 be chosen so as to contain at least 3 
physicians ? 

13. A crew contains eight men ; of these three can row only on the 
port side and two only on the starboard side. In how many ways can the 
crew be seated ? 

14. Find n when „+2C4 = llnC2- 

15. In how many ways can 18 books be divided into two groups of 6 
and 12 respectively ? Atis. isC'e. 

16. In how many ways can 12 students be divided into three groups 
of 4, 3, 5, respectively ? 

17. How many different amounts can be weighed with 1, 2, 4, 8, and 
16 gram weights ? 

18. How many sums of money can be made with 5 one-cent pieces, 
4 dimes, 2 half dollars, and 1 five-dollar bill ? 

19. In how many ways can four gentlemen and four ladies sit around 
a table so that no two gentlemen are adjacent ? Ans. 144. 



XVII, § 273] PERMUTATIONS AND COMBINATIONS 431 

20. Prove nO, + „Or_l = n+lC,-* 

21. How many dominos are there in a set numbered from double 
blank to double six ? 

22. A railway signal has three arms and each arm can take three 
different positions. How many signals can be formed ? 

23. Prove n+^Cr+i = nCr+i + 2 „0, + „a-i. 

24. How many combinations of four letters each can be made from 
the letters of the word proportion ? How many permutations ? 

Ans. 53; 758. 

25. Find the probability that in a whist hand a player will hold the 
four aces. 

26. Find the probability of drawing a face card from a pack of 52 
playing cards. 

27. If two tickets are drawn from a package of 15 marked 1, 2, •••, 15, 
what is the probability that they will both be marked with odd numbers ? 
both with even numbers ? both with numbers less than 10 ? both with 
numbers more than 10 ? 

28. To decide on partners in a game of tennis four players toss their 
rackets. The 2 "smooths" and the 2 "roughs" are to be partners. 
What are the odds against the choice being made on the first throw ? 

29. Prove that the sum of the coefficients of the odd terms of a 
binomial expansion equals the sum of the coefficients of the even terms. 

30. If n is an even integer, prove that there is a middle term in the 
expression of (x + a)" and that its coefficient is even. 

31. Provethat„Ci+2„(72 + 3„03 + ••• n„a„= w(2)"-i. 

32. Prove „Ci - 2„C3 + 3„(73 + .- (- l)"-i • w • „C„ = 0. 

* An application of this formula is the construction of Pascal's Triangle. 
(o^o by definition will be assigned the value 1.) 



oCo 








iCo 


iCi 






2C0 


aC'i 


. 2C2 




8^0 


sCi 


8C2 


3^3 


4C0 


4C1 


4C2 


4C3 













1 








2 


1 






3 


3 


1 




4 


6 


4 



404 

The formula in Ex. 20 shows that any number n+iCr is equal to the number 
just above it, i.e. nCr, plus the number nCr-i which is to the left of nCr.. Thus 
for example 4C8 = 3C3 -f 362. We can, by means of this formula in Ex. 20, 
write down the next row. It is 

1 5 10 10 5 1 
The numbers in the nth row of the table are seen to be the coefficients 
of the terms in the expansion of (z-{- a)^ (§ 273) . 



CHAPTER XVIII 
COMPLEX NUMBERS 

274. Definitions. We have already had occasions to refer 
to the so-called imaginary numbers. A number that arises as 
the result of extracting the square root or, indeed, any even root 
of a negative number is called an imaginary number. Thus 
V— 2 is an imaginary, number ; the roots of the quadratic 
equation ^624. 3 _ q^ yiz. ± 2V— 2, are imaginary numbers. 

We have hitherto avoided the use of imaginary numbers as 
far as possible. It now becomes desirable to take them defi- 
nitely into account, to learn how to work with them, and to 
gain some knowledge of their usefulness. Indeed, one of the 
primary objects of this chapter is to show that imaginary 
numbers have quite as concrete an interpretation as the real 
numbers, an interpretation which in many cases is of great 
service in the solution of concrete problems. 

The letter i is used to represent the so-called imaginary 
unit; it is by definition such that i^ = — 1. 

Numbers of the form ibj where 6 is a real number different 
from zero, are called pure imaginary numbers. 

Numbers of the form a -\- ib, where a and b are real numbers, 
are called complex numbers. 

In the complex number a -f- ib, a is called the real part and 
ib the imaginary part. In a real number the imaginary part 
is zero ; in a pure imaginary the real part is zero. A complex 
number a -f- &« is imaginary if 6 ^ 0. 

When two complex numbers differ only in the sign of the 
imaginary part they are said to be conjugate. Thus 3 + 2 t 
and 3 — 2 i are conjugate complex numbers. 

432 



XVIII, § 276] COMPLEX NUMBERS 433 

275. Assumption. We assume that complex numbers obey 
the laws of algebra given in § 41. By applying this assump- 
tion we have symbolically for the sum and difference of the 
two complex numbers a + ib and c -f id, 

a -{-lb ± (c -\- id) = a ± c -\- i(b ± d). 

That is, to add (subtract) complex numbers, add (subtract) the 
real and imaginary parts separately. 

276. The Geometric Interpretation of the Imaginary Unit. 

We now seek a geometric interpretation of the imaginary unit 
i. To this end we recall the familiar representation of the 
real numbers as directed segments on a line, o j 

together with the interpretation of multiplica- ^^co^-a a 
tion by - 1 (§ 35). To multiply a real number ^'''- ^^^ 

a by — 1 is equivalent geometrically to a rotation about the 
point through two right angles of the segment OA which 
represents a (Fig. 237). 

Now, by definition, i is such a number that i^ = — 1. To 
multiply a real number a by — 1 is then equivalent to multi- 
plying it by P, i.e. by i • i. Multiplying a real number a by i 
may, therefore, be interpreted geometrically as an operation 
which when performed twice is equivalent to a rotation about 
in the plane through two right angles ; i.e. 
to multiply a by i may be interpreted geo- 
metrically as equivalent to rotating OA about 
" ^ in the plane through one right angle. 

The number ai will then be represented by 

a segment OB equal in length to OA whose 

direction makes with that of OA an angle of 

90° (see Fig. 238). In the figure we' have also indicated the 

result of multiplying ci by i^=i 'i=—l and by i^=i • i • i= —i. 

2f 



434 MATHEMATICAL ANALYSIS [XVIII, § 276 

Multiplying by i* = i • i . i . i = 1*2 . /2 = 1 is then to be inter- 
preted as a rotation through four right angles. 

EXERCISES 

Give the conjugates of the following complex numbers : 
1. 3 + 2 I. 2. 3 - 4 1. 3. _ 5 - 3 1. 4. _ 8 + i. 

Simplify the following expressions : 
6. 2(3 + 4 i) - 4(1 - i). 7. ll^ii _ ^J^li. 

6. - 4(1 - i) -}- 6(3 - 28 i). 8. x + iy + ix + y. 

9. Prove that the sum of two conjugate complex numbers is a real 
number. 

10. Is the following statement true ? If the sum of two complex 
numbers is a real number, the complex numbers are conjugates. Explain. 

11. Prove that every even power of i is equal to either 1 or — 1. 

12. Prove that every odd power of i is equal to either i or — i. 

13. Find the value of i 4- 2 i^ + 3 i^ + 4 i*. 

14. Find the value of i^^ + i*^ + ^es _|. j69 _,. 4-44. 

277. Vectors in the Plane. We have seen that, if any real 
number a is represented by a horizontal segment directed to 
the right or left according as the number a is positive or 
negative, then the imaginary number ai may be represented 
by a vertical segment directed upward or downward accord- 
ing as a is positive or negative. This suggests the possi- 
bility of representing other complex numbers by segments 
having other directions in the plane. Such a directed seg- 
ment will represent a magnitude (the length of the segment) 
and a direction. Therefore such a segment can be used to 
represent a variety of concrete quantities that are not merely 
geometric; e.g. a force of a given magnitude and acting in 
a given direction; a velocity, meaning thereby the speed 
(magnitude) and the direction in which a body moves ; etc. 
Such quantities having both direction and magnitude are 




XVIII, §278] COMPLEX NUMBERS 435 

called vectors, and, if the directions are restricted to lie in 
the same plane, they are called plane vectors. Any plane 
vector may, then, be represented by a directed line-segment 
in the plane. 

Two vectors are said to be equal if and only if they have the 
same magnitude and the same direction. Hence,- from any 
point in the plane as initial point, a vector can be drawn equal 
to any given vector in the plane. 

278. Addition of Vectors. The addition of vectors in the 
plane proceeds according to a definition analogous to the geo- 
metric addition of directed line-segments discussed in § 35. If 
we are given two vectors AB and BC, we may 
conceive the first to represent a motion from 
Ato B and the second a motion from B to O. 
The sum of the two vectors then represents, ^ ^ 

by definition, the net result of moving from A ^^^' ^^^ 

to B and then from B to C, i.e. the motion from A to C. The 
sum of the vectors AB and ^C is then the vector AC (Fig. 
239). In symbols ^s + BO = AC. 

In other words, the sum of two vectors is the vector from the 
initial point of the first to the terminal point of the second, 
when the vectors are so placed that the initial point of the 

second coincides with the terminal point of the 

A,^^^^\/ first. From this definition it follows immediately 

^ that, if two vectors issue from the same point 

0, their sum is the diagonal, issuing from 0, of 

the parallelogram of which the two given vectors form two 

adjacent sides (Fig. 240).^ 

* If the vectors represent two forces, this shows that the sum of the vectors 
represents the resultant of the forces according to the law known as " the 
parallelogram of forces." 



436 



MATHEMATICAL ANALYSIS [XVIII, § 279 



279. The Components of a Vector. The projection of a 
vector on a given line is called its component parallel to the line. 
Thus in Fig. 241 the directed segment M1M2 is the horizontal 



»« 


r -^ 


' 


*l 


1 






i 1 




M, -M, 



Fig. 241 

component of the vector AB, and the directed segment ^1:^2 
is its vertical component. Moreover, 

vector AB — vector N^N^ + vector M^M^. 

If the horizontal and the vertical components of a vector are 
known, then the vector is known. Why ? 

280. The Complex Number x + iy and the Points in the 
Plane. Let OP (Fig. 242) be any vector issuing from 0, and 
let the horizontal vectors issuing from be represented by the 
positive and negative real numbers (and zero). We have seen 



F; 




p 


Vi 


■^ 







X J 


/ V 



Fig. 242 
that the numbers of the form ai can be represented by the ver- 
tical segments issuing from 0. Here a is a real number and i is 
a vector of unit length. The horizontal component of OP will 
then be a certain real number x, and the vertical component a 
certain pure imaginary number iy. The vector OP will then 
be equal to the sum of these two components, i.e. 
OP^x-\- iy. 



XVIII, § 280] 



COMPLEX NUMBERS 



437 



Conversely, every number of the form x -\- iy represents a 
definite vector in the plane. If its initial point is at the origin 
of a system of rectangular coordinates (with equal units on 
the two axes), its terminal point is the point (x, y). 

We have hitherto used vectors in the plane to represent the 
complex numbers. If we think of these vectors as all having 
their initial points at 0, each vector determines uniquely, and 
is uniquely determined by, its terminal point. Hence, we can 
also use a complex number to represent a point in the plane, 
viz. the number x + iy will represent the point whose rectan- 
gular coordinates are (cc, y). 

Example 1. Represent by means of vectors the complex numbers 
'2 ■+ 2i and 1 ■\-Qi. Find the vector that represents their sura. 

In Fig. 243 the vector OA represents the complex number 2 + 2 i, and 
the vector OB represents the complex number 1+6 i. The sum of these 
two complex numbers is represented by the vector OC. Why ? 







-p 




-■e 


^t 


-r, ^-!i 


^^ 71 


' it 


_, , ' 


- ===f K-"-'- = 


: :::Si " - 


irM t 


-Z^ t 


1 t 


it ^A 


t 4 


l^X- 




-0- - ^ 

























1 






























_x 


T 
















/-y 


















> 


















\\ 


















\ 


\ 


















\ 


















"^ 
















\ 


h ^i 


















^\ 


















q2I. 


















N 




Y 














^v 


















-^ \ 


















A^ 




































'' 


^ 


















B 


















• 






























































- 













Fig. 243 Fig. 244 

Example 2. Find the vector that represents (1 + *") — (2 — 3 1). 
To find this vector, find the vectors, OA and 0J5, that represent 1 + i, and 
2 — 3i, and determine OC so that OA is the diagonal through O of the 
parallelogram of which OB and OCare adjacent sides (Fig. 244). Note 
that the vector OC is equal to the vector BA. 



438 



MATHEMATICAL ANALYSIS [XVIII, § 281 



281. Equal Complex Numbers. Ifx-{-iy = 0, then x = and 
y = 0. For, if cc + iy = 0, and y=^0, we should have x/y = — i, 
which is impossible. Why ? 

If Xy + lyi = X2 + iy2, then x^ = x^ and y^ = 2/2- ^oi*) ^J trans- 
posing terms, we have (x^ — Xo) + i(yi — 2/2)= ^- Hence, we 
have Xi = X2 and y^ = 2/2. 

Thus, fwo complex numbers are equal if and only if the real part 
of the first is equal to the real part of the second, and the imaginary 
part of the first is equal to the imaginary part of the second. 
Geometrically, two complex numbers are equal if and only if 
they represent the same point. 

282. The Polar Form of a Complex Number. Connect the 
point P{x, 2/) (Fig. 245), which represents the complex number 
X 4- iy, to the origin 0. If we let {p, 6) (p > 0) be the polar 
coordinates of P(0 being the origin and OX the initial line), 

then for any position of the point F we have 

^ 'x = p cos 6, 

,y = p sin 0. 

Therefore, the complex number x -\-iy may 
be written in the form 

(2) x + iy = p(cose + zsine). (p^O.) 

This form of complex number x + iy is called the polar form. 
The angle 6 is called the angle or the argument, and the length 
p is called the absolute value * of the complex number. 

Example. Find the angle, the absolute value, and 
the polar form of the complex number 2 -}- 1 2\/3. 

Plot the complex number (Fig. 246). Now we have 
p — y/x^ + y^. Hence p = \/4 + 12 = 4. Moreover 
tan^=\/3, i.e. 6=00°. That is, the absolute 
value is 4 and the angle is 60°. Therefore the polar 
form is 4 (cos 60° + i sin 60°) . ^«- 2*6 

* Also sometimes called modulus. 



V 



(1) 



Fig. 245 



Y 


J 


Pi2+i2/3) 




/ 


2/3 




A'° 







2 


X 



XVIII, §282] COMPLEX NUMBERS 439 

EXERCISES 

In the following exercises represent by vectors the numbers in paren- 
theses, and their sum or difference as the case may be : 

1. (3 + + (-4 + 20. 4- (5-40-(-2-i). 

2. (1 +3i)-(5-60. 5. (3 + 2z)+(3+20. 

3. 7_(5 4.3 4). 6. (-4-4i)-6. 

Represent by a point each of the following complex numbers ; 

7. 3 + 5 i. 9. 6 + i. 11.-3+6 i. 

8. 3-3i. 10. -5-3i. 12. 7 + iV2. 

In the following exercises, represent by points the numbers in paren- 
theses, and their sum or difference as the case may be : 

13. (3-hO + (-4 + 2 0. 16. (5-4 0-(-2-0- 

14. (l-f-3 0-(5- 6i). 17. (3 + 2 + (3-t-2i). 

15. 7 -(5 + 3?). 18. (3 + 3 0-5. 

Find real values of x and y satisfying the equations : 

19. 2x— iy = 4:y — Q — 4i. 22. ixy + ic+?/ = 5 + 4i. 

20. X + io-y = y + 6 + 36 i. 23. a:2 + ?/2= 25 - (3 x+4 y-2b) i. 

21. {Sx + Qy+2)i — Sy-x=8. 24. ix -{- iy = 4: i -\- 5 x. 

Find the angle and the absolute value of each of the following complex 
numbers. Represent the numbers in polar form : 

25. l + iV3. 27. 1-1. 29. 3i. 31. -8i. 

26. 5 + 5i. 28 1-*-^. 30. -8. 32. 12 + 5 i. 

2 2 

33. Can the complex number x + iy, where x and y are real numbers, 
equal 7 ? 

34. Under what circumstances is the sum of two complex numbers a 
real number ? 

Change the following complex numbers from the polar form to the 
form x-}-iy : 

35. 3(cos 30° + i sin 30°) . 38. 2 V2(cos 225° + i sin 225°) . 

36. 4(cos 135° + z sin 135°). 39. 4(cos 90° + i sin 90°). 

37. cos 210° + i sin 210°. 40. 8(cos 180° + i sin 180°). 



440 



MATHEMATICAL ANALYSIS [XVIII, § 283 



283. Multiplication of Complex Numbers. Our assump- 
tion in § 275 allows us to multiply two complex numbers 
iCi -f- iz/i and X2 + iy^ as follows : 

(a?! + *2/i)(aJ2 + m) = 3^1352 4- iy 1X2 + ixiy2 + ^^l2/2 
= (aJia^2 - 2/1^2) + *(^i2/2 + X2yi)' 
If the two numbers are written in polar form, the multipli- 
cation may be performed as follows : 

^'1 + *2/i = pi(cos 61 4- i sin ^1), 
^2 + %2 = p2(cos $2 + I sin ^2)- 
By actual multiplication, we have 

(•^1 + iyi)ix2 + m) 

= P1P2 [cos 61 cos ^2 -h '' (sin Oi cos ^2 + cos Oi sin ^2) — sin Oi sin ^2] 

= P1P2 [cos (^1 4- ^2) + i sin ((9i + ^2)].* 

Therefore, the absolute value of the product of 
two complex numbers is equal to the product 
of their absolute values, and the angle of the 
product is equal to the sum of their angles. 

In Fig. 247 the points Pi and P2 represent the 
complex numbers xi + iyi and X2 + iy2 respectively. 
The point P3 represents {xi-\-iyi)(x2 + iyo)- 




Fig. 247 



284. Division of Complex Numbers. The quotient of two 
complex numbers Xi -\- iyi and X2 -f- iy^ luay be reduced to the 
form a + ib if we make the denominator real by multiplying 
both numerator and denominator by the conjugate of the 
denominator. Thus, 

Xi + iyi _ a?i + m . X2 — iy2 ^ X1X2 -h iyiXj — ixiy2 — i^yiyz 

X2 -f iyo «2 + m ^2 — iyi 



^2' + 2/2' 

* See § i:ts 



X2^ + 2/2' 



^%Vi 



x^^y^ 



XVIII, § 284] 



COMPLEX NUMBERS 



441 



If we write the two complex numbers in polar form and 
then perform the division, we have 

pi (cos $1 + I sin ^i) _ pi (cos Oi + i sin ^])(cos 0^ — ^ sin O^) 
P2 (cos $2 + i sin O2) p2 (cos O2 + i sin 62) (cos ^2 — i sin ^2) 

^ Pi [cos (^1 - O2) + i sin (^1 - ^2)1 
P2(cos2^2 + sin2|92) 

= ^^ [cos (^i - $2) + 1 sin (0, - ^2)]. 

Therefore, the absolute value of the quotient of two complex 
numbers is equal to the quotient of their absolute values, and the 
angle of the quotient is equal to the difference of their angles. 

Example 1. Find analytically and graphically the product (1 -f i) 

Solution. Analytically, 

(1 + 0(3 4. V8 i) = 3 + 3 I + V3 ^ + \/3 i^ = (3 - V3) + i (3 + V3). 

Graphically, writing the complex numbers in polar form, we have 

V2(cos 45° + i sin 45°) and 2 V3(cos 30° + 2 sin 30°). 
Therefore pi = \/2, P2 = 2 V3, Ox = 45°, 62 = 30°. 







Ir 4^ 


t^ 


£ 




1 


t 


-^rit - 


7 p^^^' 


tc-^ 


^_ 


X 














1 . . 



Fig. 248 

Hence the absolute value of the product is p\p2 =2 V6 and the angle of 
product is 75°. In Fig. 248 the points Pi, Pi, and P represent respectively 
the complex numbers 1 + i, 3 + iVS, (1 + 1)(3 + iV3). 



442 



MATHEMATICAL ANALYSIS [XVIII, § 284 



Example 2. Find analytically and graphically the quotient 

(3 + iV3)/(l + 0- 
Solution : Analytically : 

3 + iv/3 ^ 3 + iV3 1-i^ (3+ V3)-i(3-V3) ^ 



l+i 



1 + t 1 - i 





z 






^x 


c^ -t 


//^ 


gr ^_ 


jo^^ ^±: 


■-E 





Fig. 249 

Graphically^ using the results in Ex. 1, we see in Fig. 249 that the 
points Pi, P2, and P represent respectively the complex numbers 
(1 + 0, (3 + iV3), (3 + iV3)/(l + 0. 

EXERCISES 
Perform the following operations analytically and graphically : 

1. (1 + 0(2 + 2 0. _ g l-zV3 

2. (1 + i V3)(2 4-i2V3). * -3 ' 

3. (2 0(5)- T 5 + 5t 

4. (i_j-0(-2-20(- 1 + iV3). * l-» 
3 + iV3 8. n-0' - 



6. 

1 + I 2 + I 2 \/3 

Perform the following operations analytically : 
a 3 + i U. (i9 + iio + i" + ii2)7. 

l_^18t 3-29t 
10. 



7-iV2 



3 + 4 i 3 - 4i 



11. 



12. 



13. 



(2 + 1)2 (2 - 0' 

X + t Vl — X2 



16. 



V V2 / 



Vl-x^ 



17 2 + 3 ;; 3 + 2 ^ 
3 _ 4 I 3 + 4 i 



1 + i 



18. V? + 24 i. 



XVIII, § 285] COMPLEX NUMBERS 443 

285. DeMoivre*s Theorem. The result of § 283 when ap- 
plied to the product of any number of complex numbers leads 
to the following : 

I. The absolute value of the product of any number of complex 
numbers is equal to the product of their absolute values. 

II. Tlie angle of the product of any number of complex numbers 
is equal to the sum of their angles. 

If the above statements be applied to a positive integral 
power of a number, i.e. to the product of n equal factors, 
we obtain 

(3) [/3(cos 6 -\- i sin ^)]" = p"(cos nd -\- i sin n 6). 

For the special case p = 1 we obtain 

(4) (cos 6 + i sin 6)" = cos n 9 + / sin n 6. 

This relation we have just proved for the case where n is a 

positive integer. It also holds when ti is a negative integer. 

For we have 

(cos^ + isin^r= i ^cos^-tsin^ 

^ cos + 2 sin e cos2 6 -f- sin^ 6 

= cos (— 0) 4- i sin (— 0), 

and hence 

(cos + i sin 6)-'' = [cos (— 0) -f i sin (— 0)^ 

= cos (— p 6) 4- i sin ( — p 0). 

Further, \i n — 1/g, where g is a positive or negative integer, 
we have, by what precedes, 

(5) (cos + i sin Of = [Uo^- + i sin-Vl^ 

and hence 

(6) (cos ^ + i sin Oy = (^cos - -f- / sin ^\ = cos ^ ^ + / sin ^ 9. 



6 

13= COS-+ isin-, 

q q 



444 MATHEMATICAL ANALYSIS [XVIII, § 285 

This shows that relation (4) is valid for all rational values ofn. 
It should be noted, of course, that relation (5) states merely 
that a certain q^^ root of cos 6 + i sin 6 is cos {O/q) -f- i sin {6/q) 
and that a similar statement applies to relation (6). The fact 
expressed by (4) is known as De Moivre's theorem.* 

286. "Powers and Roots of Numbers. De Moivre's theo- 
rem often enables us to compute an integral power of a complex 
number without difficulty, as the following example will show. 

Example 1. Find the value of (2-{-2i)^. The polar form of this 
number is 2 \/2(cos 46'^ + i sin 45") . Hence 

(2 + 2 0^= (2\/2)6(cos225° + i sin 225°) 

= 128V2(---- 7:^=- 128 -128 I. 

V V2 V2y 
To find the nth roots of a number requires special methods. 

Example 2. Find the 5th roots of 2 + 2 i. 
Here as in Ex. 1 we may write 

2 + 2 i = 2V2(cos 45° + i sin 45°) 
and hence (2 +2 0^=(2V2)^(cos9° + isin9°). 

But this is not the only number whose fifth power is 2 +2 i. For we 
may write 2 + 2i = 2v'2[cos(45° + k 360°)+ i sin(45° + fc 360°)], where 
k is any integer. That is to say, 

(2 + 2i)^=(2\/2)^[cos(9°+ ^•72°)+ isin(9° + Ar72°)]. 



For the values A; = 0, 1, 2, 3, 4 we get the five numbers 
(7) 



(2\/2)^"(cos9° + isin9°), (2 V2)^ (cos 225° + i sin 225°), 



(2\/2)^^(cos81° + i sin 81°), (2V2)^(cos297° + i sin 297°). 
[ (2 V2)^(cos 153° + i sin 153°), 

The succeeding values of k (i.e. A = 5, 6, •••) evidently give numbers 
equal to the preceding respectively. Each of the five numbers is a fifth 
root of 2 + 2 i ; they are all different. 

♦Abraham de Moivre (lfi67-1754), a mathematician of French descent 
who lived most of his life in England. 



XVIII, § 286] COMPLEX NUMBERS 445 

The general formulation of the problem of finding the nth 
root of a number z = /o(cos 6 -\-i sin 9) is as follows. The most 
general form for z is 

z = p[Gos{e + k 360°) 4- i sin((9 + k 360°)], 

where k is an integer. 

This gives, by De Moivre's theorem, 

1 1 
z'-==p' 



(9 + A: 360° , . . 6> + A; 360 

cos --^ h t sm —^ 



The n values A: = 0, 1, 2, —, n — 1 give n different values for 
2^/" and no more values are possible. Why? Here pV« means 
the numerical nth. root of the positive number p. We have 
then: Every com2i>lex number (^0) has just n nth roots. These 
n roots all have the same absolute value ; their angles may be 
arranged in order in such a way that every two successive 
ones differ by 360°/7i. 

EXERCISES 
By using De Moivre's theorem find the indicated powers, roots, and 
products. 

1. (4 4-i4V2)6. 4. (3 + iV3)io. 

2. (cos 10° + i sin 10°)9. 5. ( - 1 - WS)^. 

3. (l+iV^. 6. (-2+2«)*. 

7. [3(cosl5° + isinl5°)]i5. 

8. [2(cos 20° + I sin 20°)][3(cos 70° + i sin 70°)]. 

9. [2 4-2i][V8 + t]. 

10. (3-3 0(-l + iV3). 

11. \/4 + i4>/2. 
12- V3 + tV3. 
13. </_4+4z. 



15. 


V- l-iV3. 


16. 


v^cos 45° + i sin 45^. 


17. 


V2Ti. 


18. 


The cube roots of 1. 


19. 


V:^. 


20. 


V2^. 



14. \/8(cos6U° + isin60^). 

21. Prove that tlie n nth roots of a given number z are represented by 
the vertices of a regular polygon of n sides whose center is at the origin. 



446 MATHEMATICAL ANALYSIS [XVIII, § 287 

287. Applications in Trigonometry. De Moivre's theorem 
may be used to advantage in certain trigonometric problems. 

I. To express cos nO and sin nd in terms of cos and sin 6. 
' We have the relation 

cos nO + i sin nO = (cos $ -{- i sin Oy 

= cos" O-hn- i cos"-i ^ sin ^ -f ^K^^ — ^) ^-2 cos"-2 sin^ ^ + ... 

A 

If in this relation we equate the real and the imaginary 
parts we get the expressions desired. 

Example 1. Express cos 6 d and sin 6 ^ in terms of cos 6 and sin d. 
The above method yields in this case : 

cos 6 ^ + i sin 6 ^ := (cos 6 ■{■ i sin 0)'^ 
= cos6 e -\-Q icos^ ^sin ^ - 15 cos* 0sin2 ^ - 20 1 cos^ 6 sin' 6 + 16cos2 ^sin* d 
+ 6 I cos 9 sin° 6 — sin^ d. 

Equating the real parts, we have 

cos 6 £> = coss e — 15 cos* e sin2^ + 15 cos'^ sin* d — sin^ 9. 

Equating the imaginary parts we get (after dividing by i) 
sin 6 ^ = 6 cos5 d^inO — 20 cos^ d sin^ ^ + 6 cos ^ sin^ d. 

II. To express cos"" 6 and sin"" 6 in terms of sines and cosines 
of multiples of 0. If we place w = cos + i sin 0, we have 

u^ = cos kO 4- i sin kO, u~^ = cos kO — i sin kO. 

Adding and subtracting these equations, we have 
.r,. [ w* + w'' = 2 cos k$, 

[ u^ — n -* = 2 4 sin kd, 

for any integral value of k. 

In particular when /c = 1, we have 

2 cos 6 = u -^ u~^, 2 i sin $ = u — ^^~^ 

It follows that 
2^ cos" e= {u -|-ifc-i)«= w"4- nu-'^-{- ^^^'~^\ ''-*-\ \-na-^''-^^+u'\ 



XVIII, § 287] COMPLEX NUMBERS 447 

The fact that the coefficients in the binomial expansion are 
equal in pairs makes it always possible to group the terms as 

2'^ cos"^ = {w H- ?r") -h n(ifc'^-2 -f ir(«-2))+ .... 

But the terms in parentheses on the right are equal respec- 
tively to 2 cos ?i^, 2 cos {n — 2)0, •••. The following examples 
will make the method clear. 

Example 2. Express cos* in terms of cosines of multiples of 6. 
We set 

= w4 + 4m2_|.6 + 4 u-^ + M-* 
= (m* + M-*) + 4(m2 _|_ u-2) -f 6 
= 2 cos 4 ^ + 4 . 2 cos 2 + 6. 

Dividing both members by 2* we obtain the desired result 
cos* e = I (cos 4 ^ + 4 cos 2 ^ + 3) . 

Example 3. Express sin^ 6 in terms of multiples of the angle d. 
We set 

25 1*5 sin^ e = {u— u-^y 
or 

32 i sin^ 6 = u^ — bu^ -\-10 u— Id ir^ + 5 m-^ — u-^ 

= («5 _ |<-5) _ 5(1(3 _ «-3) 4. 10(w _ M-1) 

= 2 I sin 5 ^ - 5 • 2 i sin 3 ^ 4- 10 . 2 i sin d. 
Whence 

sinS d = ^^ (sin 5 ^ — 5 sin 3 ^ + 10 sin ^) . 

EXERCISES 
Express each of the following in terms of cos d and sin 0. 

1. cos 2 and sin 2 ^. 3. cos 4 ^ and sin 4 ^. 

2. cos 3 e and sin 3 ^. 4. cos 5 ^ and sin 5 0. 

6. Show that tan 4 g ^ ^ <^^^ ^ (^ - ^^"' ^) . 
1-6 tan2 + tan* ^ 

6. Find tan 5 ^ in terms of tan 0. 

Express each of the following in terms of sines and cosines of multi- 
ples of ^ : 

7. sin8 0. 9. sin* 0. 11. cos^ 0. 

8. cos8^. 10. cos«^. 12. sin6^. 



418 MATHEMATICAL ANALYSIS [XVIII, § 287 

MISCELLANEOUS EXERCISES 

Solve the following equations and illustrate the results graphically. 

1. x3 - 1 = 0. 3. x5 - 32 = 0. 5. cc8 - 1 = 0. 

2. x3 + 1 = 0. 4. a;6 - 1 = 0. 6. a:^ + 1 = 0. 

7. Prove that 

cos nd = I [cos d + I sin ^]" + ^[cos 6 — i sin ^J". 

8. Prove that 

i sin nd = I [cos 6 -\- i sin ^]" — i [cos — i sin ^J**. 

9. Prove that 

/I -f sin ^ + t cos ^X'* ,, /,N , • • /T „N 

{ '—- — =cos(lmr—nd)+ismUn'ir — nd). 

10. Prove that the product of the n nth roots of 1 is 1, if n is odd, and 
— 1 if n is even. 

11. Prove that the sum of the n nth roots of any number is 0. 

12. Complete the discussions in § 287 to derive the following formulas. 

L (a) cos;i^=cos"0— ^^^^i^~!-)cos"-'^^sin2^ 
^ ^ 2! 

, n(n— l)(n — 2)(n — 3) „ a ^ ■ a ^ . 

_l — V ZA zs 1 cos "-4 d sm^ 6 -\- •••. 

4 1 

(6) sin nd=:n cos'»-i ^ sin ^ - ^(^ - ^) ( ^.ZL^ cos^-s q gin^ ^ 

o ! 

, nCn- l)(n - 2)(n-3)(n-4) „ ,^ . ,^ 

+ ~^^ ^-^ — — ^ cos"-fi e sin^ ^4- . . •. 

5 ! 

II. (a) cos«^=-l- rcosn^+ncos(n-2)^+*-^— ^cos(n-4)^+-"l. 

n 

(b) sin»^ =^-=^1 cosn^- ncos(n - 2)^ +^^^=-^cos (n-4)^+ •••], 
if n is even ; but 



sin"/? = i=J-) ^ rsinn^-nsinrn-2)^4- ^^^~-^^ sii 
if n is odd. 



fsin «^-n sin(n-2)^ + ^^^ ^^ sin (n - 4)^+ ...], 



CHAPTER XIX 



THE GENERAL POLYNOMIAL FUNCTION 
THE THEORY OF EQUATIONS 

288. The General Polynomial Function of Degree n. The 

general polynomial of degree n, 

f{x) = a^x"" 4- a^-ix""-^ + a,^-2X^-^ H \- aiX -{- a^ (a„ :^ 0), 

has already been defined (§ 255). We have already dis- 
cussed in some detail special cases when the degree of f(x) is 
1, 2, 3, (Chapters III, IV, V). For these cases we proved that 
the function is always continuous, and we learned how to find 
the slope of the graph of the function at any point. It is our 
present purpose to extend these results and methods to a func- 
tion represented by a polynomial of any degree. 



ax^ + a^-iX""-^ H h aiK + otr 



289. The Slope of the Graph of /(x). 
the slope of the graph of the equation 

(1) y=n^) 

at any point Pi{xi, y^ of this graph, 
we first find the slope A^z/Aic of the 
secant P^Q (Fig. 250) joining the 
point Pi to any other point Q(flJi+ Aa?, 
2/i + Ay) on the graph. To this end 
we must first calculate the value 
of Ay in terms of Xi and Aoj. We 
have 

2o 449 



Continuity. To find 




Fig. 250 



450 MATHEMATICAL ANALYSIS [XIX, § 289 

= a,(a^i + ^xy + a«_i(a:i + Axf-'^ + - +ai(xi + Ax)-\-aQ. 
Vi =/W= «„a;i" + a„_iaji«-i + .•• + a.x^ + a^. 
By subtraction and proper grouping of terms we find 

(2) Ay=f(x,-{.Ax)-f(x{) 

= alix, + Axy- a^r] + a„_i[(a;i + Axy-^-x^^^-] 

H h«i[(^i + Aa;) — ajj. 

Each of the terms of this expression is of the form 

(3) al(x, + Axy-x,'^l 

and the whole expression is equal to the sum of all terms ob- 
tained from (3) by letting k take on the values k = n,n^l, —, 1. 
Expanding the first term in the brackets, we obtain 
a,l{x^ + Axy - X,''] 

= «,[.T/+A;a;i*-iAa;+ ^fc^a;i*-2Aa;2 + ... + Aa;*- a^i*] 

A 

^alkx^-^^^-~-^x^-''b.x + - + Aa;*-i]Aa?. 

Z 

It is clear from this expression that for every value of k 
the expression (3) has A« as a factor. Moreover the expres- 
sion (2) for A?/ is the sum of such terms as (3) for different 
values of k ; and, since each of these terms has the factor Ax', 
their sum has the factor Aa;. Hence, if we divide Ay by Aa;, 
we have for the slope Ay/Aa; of PiQ, the expression 

Aa/ u 

-f terms with higher powers of Aa;] k=.n. 

+ a,_j[(n - l)a;i"-2 + (^^ - l)(n - 2) ^^,^_3^^ 

4- terms with higher powers of Aa?] k = n — \. 



f a2[2 a^i + Aa;] A: = 2. 

-h tti A: = L 



XIX, § 290] GENERAL POLYNOMIAL FUNCTION 451 

The slope m of the graph is the limit approached by Ay/Aa; 
as Aic approaches zero (i.e. as Q approaches Pi along the curve). 
This gives finally 

(4) m = ?ia„iCi'*-^ +(n — l)an-i^i''~^ H -f ^ ^2^n + «i 

[Note that for the values w = 3 and w = 2 this reduces to the expres- 
sions previously derived for the cubic and quadratic functions. ] 

Moreover, it follovt^s from the remark above, concerning the 
fact that Aa; is a factor of A?/, that as Ax approaches zero, Ay 
approaches zero also. But this proves that f{x) is continuous 
for every value of x. We have then the theorem : 

Every polynomial f{x) is a continuous function of x. 

290. The Derived Function. In previous cases where we 
have considered the slope of a curve y =f(x) we have always 
considered its value at some given point Pi on the curve. As 
the point Pj moves along the curve, however, the value of the 
slope in general changes. In other words, the slope itself may 
be considered as a function of a;. This function is called the 
derived function or the derivative of f(;x). If the original 
function is denoted by f{x), the derived function is denoted 
hj f'{x). In case of the polynomial f{x) considered in the last 
article the derived function f\x) is obtained from the expres- 
sion for the slope m by letting the given value Xi become the 

variable x, i.e. if f(x) = a^x'' + cin-i^'"'^ H h ^i^^ + «05 ^® ^^^® 

the derived function 

(5) f\x)= na,x^~^+ (n - l)a„_ia;»-2+ ... + a^. 

The derived function of any polynomial is readily written 
down from the following consideration. The derivative of any 
term a^x'' is Zca^ic*"^ ; i.e, it is obtained by multiplying the term 
by the exponent ofx and reducing the exponent of x byl. Thus 
the derivative of a^ is 3 x^, of 10 a;^ is 20 x. The above expres- 
sion for/' (a;) shows that the derivative of a polynomial is the 



452 MATHEMATICAL ANALYSIS [XIX, § 290 

sum of the derivatives of its terms. Thus the derivative of 
6 a;^ — 3 ic* + 7 a;2 — 1 may be written down at once ; it is equal 
to 35 a^ — 12 a;3 _j_ 14 ^.^ Observe that the derivative of a con- 
stant is 0. 

The relation between the derived function f{x) and the slope 
of the graph at any point, is expressed as follows : 

The slope of the graph of the curve y =f{x) at the point x = Xi 
is equal to the value of the derived function for x=Xi^ i.e. m=f'{x^. 

Further, since the derived function of a polynomial is a 
polynomial, it follows from the theorem at the end of the last 
article, that the derived function of a polynomial f{x) is a con- 
tinuous function of x. 

EXERCISES] 

Find/Cx) when 

1. /(a;) = x3 + 4 x2 - 6 a; + 3. 

2. /(a;) = 5a:6-4a;8 + 6a;2 + 2a; + l. 
8. /(x) = 7 a:7 - 4x3 + 2 X + 19. 

4. /(x)=3x6-4x* + 2x8 + 3x2+ 1. 

6. Find the equation of the tangent to 2/ = 4 x* ~ 3 x + 1 at (1, 2). 
6. Find the equation of the tangent to y = x^ — 5x2 + 2 at the point 
(1,-2). 

291. The Graph of a Polynomial S{x). In drawing the 
graph of a given polynomial of degree greater than 3, we may 
proceed as in the cases of polynomials of degrees 2 and 3. 
There are two general theorems to aid us : 

(1) The graph of any polynomial is a continuous curve ; in 
particular, the value of y does not become infinite except when 
X becomes infinite. 

(2) The tangent to the graph at any point P turns continu- 
ously as P moves along the curve ; i.e. the curve has no sharp 
corners and the tangent is nowhere vertical. (Why ?) 

We found in discussing the graphs of cubic functions that 



XIX, § 291] GENERAL POLYNOMIAL FUNCTION 453 



the values of x for wMcli the slope is zero were particularly- 
helpful, ill view of the fact that they gave us, in general, the 
turning points (maxima and minima) of the graph. Let us 
apply these principles to an example. 

Example. Draw the graph ofy=f(x)= -|(3 a^— 4 x^ - 12 «2 + 3). 

Weliave f(x) = i(x^- x^-2x)= 4x(a;- 2)(x + 1). 
Hence/(x) = when a; = 0, 2, — 1. 

We require next a table of correspond- 
ing values of x and y. Here synthetic 
division is often convenient. Thus, to find 
/(ic) when a; = 2, we write 

3 -4 -12 3[2 

6 4-16-32 



3 2-8-161 


-29 = 32/. 


Hence y = — 9| when x : 


= 2. 


When a; = 3, we have 




3 - 4 - 12 
9 15 9 


3|3 

27 



3 6 3 9 30 = 32/. 

Hence y = 10 when a; = 3. 



:::::i^-::"t: 




22 


^__.5. 


==-========;== 


::::3::^:::i=::: 






;::::i:::|i::i: 


^^^ 



Fig. 251 



We may note that since all the partial 
results 3, 5, 3, 9, 30 are positive, any value of a; > 3 will give values of y 
greater than 10. 

Finding the values of y for other values of x, we have the following 
table: 



x = 


-2 


-1 





1 


2 


3 


y = 


111- 


-1 


1 


-3^ 


-H 


10 


w = 


















We have also indicated in the table the values of x for which m is zero. 
These data give us the graph exhibited in Fig, 251. This example sug- 
gests certain other general theorems regarding the graph of a polynomial, 
which are discussed in the following articles. 



454 MATHEMATICAL ANALYSIS PCIX, § 292 

292. The Value of a Polynomial for Numerically Large 
Values of x. In the example of the last article we saw that 
for all values of x>S, the values oif(x) were greater than 10 ; 
in fact, the nature of the synthetic division showed that as x 
increased indefinitely from x = 3, the value of f{x) increased 
indefinitely. Any polynomial f(x) of degree n with real coefii- 
cients (§ 288) may be written in the form 

f{x) = aAl +f^^i=l^' + ^^=?^' + - +-^^1 
[_ \ ttna?" anX"" anX'^Jj 

=a„x.[l+(o„_,l+c„.,l+...+c.i)]. 

Since the absolute value of a sum is equal to or less than 
the sum of the absolute values of its terms (§ 35), we have, 






X'^l 

11. 
< 



1 

Gn-1- 
X 



X'^l 



1 

Cn — 



i|(|c,.i| + |c„_2| + - + |eo|)<-^, (kl>l). 

x\ \x\ 



where c is a positive number independent of x. Hence, if | a; | >c, 
the value of the expression in square brackets above is cer- 
tainly positive. Therefore for sufficiently large values |a;|, 
the sign oif{x) is the same as the sign of a„ic'». 

If an is positive and x becomes positively infinite, f(x) is 
positive. If a„ is positive and x becomes negatively infinite, 
f{x) is positive if n is even, and negative if n is odd. If a^ 
is negative and x becomes positively infinite, f(x) is negative. 
If a„ is negative and x becomes negatively infinite, f(x) is 
negative if n is even, and positive if n is odd. 

As X increases indefinitely in absolute value, the value of f(x) 
increases indefinitely in absolute value. For sufficiently large 
values of\x\, the sign off{x)is the same as the sign, of anX"". 



XIX, § 293] GENERAL POLYNOMIAL FUNCTION 455 

In particular, this leads us to the following theorems. 

Iff{x) is a polynomial of even degree, the infinite branches of 
the graph ofy = f{x) are either both above the x-axis or both below 
the X-axis (according as a„ is positive or negative). 

If f(x) is a polynomial of odd degree, the infinite branches of 
the graph of y =^ f(x) are on opposite sides of the x-axis (below 
the a;-axis on the left and above the a>axis on the right, if a,^>0 ; 
above the ic-axis on the left and below on the right, if a^<0). 

From these theorems and from the continuity of the function 
f(x) we derive the following corollary. 

The graph of a polynomial f{x) of odd degree with real coeffi- 
cients must cross the x-axis at least once and, if it crosses more 
than once, it must cross it an odd number of times. The graph 
of a polynomial of even degree with i-eal coefficients either does 
not cross the x-axis at all or it crosses it an even rmmber of times. 

293. The Zeros of a Polynomial /(x). The Roots of the 
Equation /(x) = 0. A value of x for which f(x) — is called 
a zero of f{x) ; i.e. if f(b)= 0, then 6 is a zero of f(x). The 
zeros of f(x) are, therefore, the values of x which satisfy the 
equation/ (a;) = 0. The zeros of /(a;) are called the roots of the 
equation f(x) = 0. The factor theorem (§ 261) tells us that if 
a is a zero of f{x), then a; — a is a factor of f(x). Since a 
polynomial of degree n cannot have more than n distinct fac- 
tors of degree one, we may state the following theorem. 

A polynomial f(x) of degree n cannot have more thann dis- 
tinct zeros. 

Since at the turning points of f(x) the slope is always 
zero, it follows from the fact that the derived function is of 
degree n — 1 that a polynomial f{x) of degree n cannot have 
Tnore than n — 1 turning points {maxima andminimxi). 



456 MATHEMATICAL ANALYSIS [XIX, § 294 

294, The Number of Roots of /(x) = 0. We have seen that 
every quadratic equation has two roots which may be real or 
imaginary and which may be equal. We have also seen that 
every cubic equation f(x) = 0, whose coefficients are real, has 
at least one real root. If this root be ri, we may write (§ 261), 
f{x) = (x — rj Qix), where Q(x) is a polynomial of degree 2. 
The latter has two zeros, real or imaginary, so that any cubic 
function with real coefficients may be resolved into 3 linear 
factors, ^(^) ^ ^^(^ _ ^^)(^ _ ,,^)(^ _ ^^y 

It may be proved that any polynomial {no matter whether the 
coefficients are real or imaginary) has at least one zero (real or 
imaginary). This statement is called the fundamental theorem 
of algebra. We shall accept it as valid without proof, since 
its proof is too difficult for an elementary course.* From this 
theorem it is easy to prove the following theorem : 

Any polynomial f(x) of degree n may be resolved into n linear 
factors. 

Proof : By the fundamental theorem, f(x) has one zero. 
Denote it by r^. The factor theorem then gives 

f{x)^{x--r,)Q,{x), 

where Qi is a polynomial of degree n — 1. By the funda- 
mental theorem, Qi{x) has a zero r2. Hence 

q,ix) = (oj - ro)Q2(a;), or f{x) = (a; - r;){x - r.,)Q2{x). 

Again, Q2(x) is a polynomial of degree n — 2. If n > 2, Qg 
has a zero, say r^, which leads to the expression 

f{x) = {x- ri) (x - rg) {x - rg) Qz{x), 

where Qi(x) is a polynomial of degree n — S. Continuing this 

* This theorem was first proved by Gauss in 1797 (published 1799) when he 
was 18 years old. For proof see Fine, College Alyelfra, p. 588. 



XIX, § 294] GENERAL POLYNOMIAL FUNCTION 457 

process we find 

/(a;) = (a; - ri)(a; ^ rj) "• (aJ - r„)Q„, 
where Qn is a constant which evidently must be a„ if f(x) is 
a^Xn + ••• + Oq. We have then finally 

f(x) = a^(x - ri)(x - r.2) - (x - r„). 

Each of the numbers Vi, r2, •••, r„ is a root of the equation 
f(x) = 0. This proves the theorem just stated. 

Moreover, no number different from Vi, rg, •••, r„ can be a root 
of this equation. For suppose s were such a number, then we 
should have /(s) = a„(s — ri){s — r2) ••• (s — r„). Since each 
of these factors is under the hypothesis different from zero, 
the product /(s) is different from zero. Some of the num- 
bers ri, r2, •••, r„ may be equal, however. This possibility 
leads to the following definitions. If f{x) is exactly divisi- 
ble by a; — r but not by (a; — r)^, then r is called a simple 
root of f(x) = 0. If f(x) is exactly divisible by (x — ry but 
not by (a; — r)^ then r is called a dovble root of /(x) — 0. If 
f{x) is exactly divisible by (a; — r)* but not by (x — r)^^ then 
r is called a A:-/oM root, or a root 0/ order k. A root of order 
greater than one is called a multiple root. If /(a;) represents 
a polynomial, the equation f(x) = is called an algebraic 
equation. Then we may state the last theorem as follows : 

Every algebraic equation of degree n has n roots and no more, 
if each root of order k is counted as k roots.* 

EXERCISES 

1. Is 1 a zero of the polynomial x'' — Sx^ + 2z* — x-\-S? 

2. Is 2 a zero of the polynomial a;* — 16 ? 

3. Is 3 a root of the equation x^ + Zx'^-{-x-Bz=0? 

* It is logically necessary to note the fact that, if exactly k of the roots 
^i> ^2» — equal r, f{x) is divisible by (k — r)* but not by (a; — r)*+i. Why? 



458 MATHEMATICAL ANALYSIS [XIX, § 294 

4. Find k so that a; =1 is a root of the equation a;^ + A^x^ — x -f 1 = 0. 
6. Pind k so that 2 is a root of the equation x^ + x^ — Ax + 3 = 0. 

6. How many roots has the equation x'^ + x8 + x + 3 = 0? How 
many of these roots are positive ? 

7. How many roots has the equation x^— 2 x* + x^ — 3 x2+2 x— 1=0 ? 
How many of these roots are negative ? 

8. Find graphically the real zeros of the functions 

(a) xs - X. (6) x3 + 2 X - 1. (c) x^ + 3 x + 2. (d) x^ - x2 - 6 x + 8. 
Draw the graph of each of the following functions : 

9. t/ = t^3j [3 x* - 4 x8 - 24x2 + 48 X + 13]. 

10. y = i^j [3 x* + 8x8 - 6 x2 - 24 X - 12]. 

11. ?/ = ^Jj [3 X* + 4 x8 - 12 x2 + 24]. 

12. 2/ = 2 x* - 14 x3 + 29 x2 - 12 X + 3. 

13. Prove, without assuming the fundamental theorem of algebra, 
that every algebraic equation of odd degree with real coefficients has at 
least one real root. 

295. Successive Derivatives. The derived function of a 

polynomial f(x) of degree n is a polynomial f'{x) of degree 

n — 1. The derivative of f{x) is a polynomial of degree 

71 — 2, is denoted hy f"(x), and is called the second derivative 

of f{x). Similarly, the derivative of f"{x) is called the third 

derivative of f(x) and is denoted by f"'{x). Similarly, the 

fourth, fifth, etc. derivatives may be found. The nth derivative 

of a polynomial of degree n is evidently a constant. 

Thus, if /(x)=x*-3x3-7x + 2, we have /'(x) =4x8-9x2 - 7, 
/"(x) = 12x2- 18x, /'"(x) = 24x- 18, /i^(x) = 24. 

296. Taylor's Theorem. The following formula is known 
as Taylofs theorem: 

(6) /(x) = /(a) + /'(a)(x - a) +^ (x - af + ... 

+m(x-ar. 

n I 
This formula enables us to express any polynomial in a? as a 
polynomial in a? — a, where a is any constant. 



XIX, § 296] GENERAL POLYNOMIAL FUNCTION 459 

For example, if we have/(ic)= a:^ _ 4 ^ + 2 and desire to express /(x) 
in terms of x + 1, we first find /'(x) = 3 x^ - 4 ; /"(x) = 6 x ; /'"(x) = 6. 
The coefficients in the above formula are, for a = — 1, 

/(-l)=5, /'(-1) = -1, /"(-l) = -6, /'"(-1) = 6. 

Therefore we have, from (6), 

x3 _ 3x + 4 = 5 -(X + 1)- 3(x + l)H(x + 1)3. 

Proof. We have seen in § 290 that the derivative of a;* is 
kx^~^. Likewise the derivative of {x — a)* is k(x — ay~\ For 
if y z=i{x — ay, we have, as in § 289, 

y + Ay ={x-\- Ax— a)*=[(a;— a)+ Ace]* 

= (x — a)*+ k(x — ay~'^Ax + terms with a factor Aaj^. 
Hence 

-^ = k{x — a)^^ + terms with a factor Ax, 

Ax 

and the limit of Ay /Ax is obviously k(x — tt)*~^ 
Let us now set 

(7) /(aj)=4, + A(«- a) 4-^2(0; -a)2+ - + J.,(a;- a)* +-. 
We then have, by taking successive derivatives of both sides, 
f(x) = ^li + 2 A2{x - a) -f ... + kA,{x - ay-' + ..., 
r(x) = 2 ^2+ - + A;(A^ - l)A,(x - af-' + ..-, 

/(*)(a;) = Zc ! ^;t + terms containing {x — a) as a factor. 

These relations must all be true for all values of x ; hence 
they must hold when x = a. But this gives 

f{a)=A„ f'{a) = Au /»=2^2,-, /*(«) = A: ! ^„ -. 

Hence 

By substituting these values in (7) above, we obtain Taylor's 



460 MATHEMATICAL ANALYSIS [XIX, § 296 

theorem as given in relation (6). Another form of Taylor's 
theorem is obtained by replacing a; by a; + a in relation (6). 
This gives 

(8) /(x + a) = f(a) +/'(fl)x +^-^x^ + ... +i^x^. 

EXERCISES 

1. Write down the successive derivatives of the following polynomials : 
(a) x8 + 4a;2_i2a;+l7. 

(6) 2x*-3a:8 + 8x2-14x + 18. 

(c) a;5 + 2 a;- 1. 

(d) 1 -3x +4x2 + 5x8. 

2. Prove that the nth derivative of a^x" + an-ix"-i + ••• + aix + oo is 
equal to OnTi I . 

3. Expand each of the following by Taylor's theorem : 
(a) x3 + 4 x2 — 12 X + 17 in terms of x - 1. 

(6) 2 X* - 3 x3 + 8 x2 - 14 X + 8 in terms of x - 2. 

(c) x^ + 2 X — 1 in terms of x + 1 . 

Id) 1 — 3 X + 4 x2 + 5 x3 in terms of x + 2. 

4. By relation (8) in § 296 express each of the following as a polyno- 
mial in X : 

(a) /(x-l)if/(x)=x5 + 4x2-12x + 17. 

(h) /(x-2) if/(x)=2x4-3x8 + 8x2-14x + 8. 

(c) /(x + 1) if/(x)=x6 + 2x-l. 

W /(x + 2)if/(x) = l-3x + 4x2 + 6x8. 

297. Multiple Roots. If we apply Taylor's theorem succes- 
sively to/(ic) and/'(ic), we obtain 

(9)/(») = 

(10) /'(») = 

f\a) +f"{a){x - a)+i^ {x - ay+^^ix - ay+ ..... 



XIX, § 297] GENERAL POLYNOMIAL FUNCTION 461 



If /(a)= 0, the first relation shows that x—a is a factor of /(a;) ; 
this constitutes a new proof of the factor theorem. If /(a;) 13 
divisible by a; — a but not by {x — ay, it follows that /(a) =n d 
and that f'{a) ^ 0. Hence by (10), or by the factor theor3n_, 
fix) is not divisible by a; — a. If f{x) is divisible by (x — a)- 
but not by (x — df, we have, from (9), /(a)=0, /'(a)=0, 
f"{a)=^0. We then conclude from (10) that if a is a double 
root of f{x) — 0, it is a simple root of f'{x) = 0. In general, if 
f(x) is divisible b}^ (x — a)* but not by (x— a)*+\ relation (9) 
shows that /(a) =/'(«) =/''(«) =- =/'-'(«) =0; /^a) =?t 0. 
Hence, by (10), f'(x) is divisible by {x — a)*~^ but not by 
(a? — a) *. This leads to the following theorem. 

A simple root of f(x)= is not a root of f\x)=0. A double 
root of f(x)= is a simple root of f'{x)= 0. In geyieral, a root 
of order k off(x)= is a root of order k — 1 off'(x)= 0. 

The following corollary of this theorem is evidently true. 

Any multiple root off(x)==0 is also a root off\x)=0. If 
f(x) andf'(x) have no common factor, f(x) = has no multiple 
roots. If(f> {x) is the H. C. F. off(x) and f'{x), the roots of 
<ft (x)=0 are the multiple roots off(x) = 0. 

Example 1. Examine for multiple roots the equation 

/(x) = x3 4- x2 - 10 x + 8 = 0. 

We have /' {x) =Sx'^ + 2x- 10. To find the H. C. F. of /(x) and f\x) 
we proceed as in § 259 : 

8a8-f3a:2-30x+24 

3x8 + 2x2- lOx 

x + 1 



3x 



X2- 


20 X 


+ 24 


8x2- 


60 X 


+ 72 


8x2 + 


2x 


-10 



-62X + 82 



3x2 


+ 2x- 


10 


186x2 


+ 124x 


-620 


-186x2 


+ 246X 






370 X 


-620 



It is now clear that /(x) and /'(x) have no common factor, 
we conclude that/(x)= has no multiple roots. 



Hence 



462 



MATHEMATICAL ANALYSIS [XIX, § 297 



Example 2. Examine for multiple roots the equation 

/(x)=x*-2x3 + 2x-l = 0. 
We have /'(x) = 4 x^ - 6 x^ + 2. 



2x*- 
2x*- 


-4x8 + 4x 
-3x8+ a; 


-2 




- x8 + 3x 
2x8-6x 
2x8-3x2 


-2 

+ 4 
+ 1 




3x2-6x 

x2-2x 


+ 3 
+ 1 



X 


2x3-3x2 + 1 

2x3-4x2 + 2x 




x2 _ 2 X + 1 
a;2 _ 2 X + 1 


2x4-1 






Hence (x — 1)2 is the H. C. F. of /(x) and /'(x), i.e. x = 1 is a triple root 
of / (x) = 0. The fourth root of /(x) = is x = — 1. How is it obtained ? 

EXERCISES 

1. Examine for multiple roots each of the following equations : 

(a) x3-3x2-24x-28=0. (6) x* + xS + 1 = 0. 

(c) x5-7x3-2x2 + 12x + 8 = 0. 

Id) x5 + X* - 9 x8 - 5 x2 + 16 X + 12 = 0. 

(e) x4-6x8 + 12x2-10x + 3 = 0. 

(/) x8-3x6 + 6x8-3x2-3x + 2 = 0. 

2. Prove that the graph of y =/(x) is tangent to the x-axis at a point 
representing a multiple root. 

3. Prove that the graph of 2/=/(x) crosses or does not cross the 
X-axis at a point representing a multiple root according as the order of 
the root is odd or even. [Hikt : Use Taylor's theorem.] 

4. Prove that a root of order k of /(x) = is a simple root of 
/*-i(x)=0. 

298. Complex Roots. If a -\- hi (a, 6 real numbers, i2= — 1) 
is a root of an algebraic equation /(«) = with real coefficients, 
then a — bi is also a root of the same equation, 

By hypothesis a -f- 6i is a root of the equation 

f{x) = anx- + a„_iaJ"-^+ ••• + ao = 0, 
i.e. f{a + 60 = a^ia + biY+ a„_i(a + bif-^^ + ••• -f Oo = 0. 
If each of the terms in the preceding expression be expanded 



XIX, § 298] GENERAL POLYNOMIAL FUNCTION 463 

by the binomial theorem, the powers of i reduced to their 
lowest terms (i^ = — 1, ^s = — i, etc.), and terms collected, we 
obtain ^, , t-\ n , r\' 

where P represents the sum of the terms independent of i ajid 
Q is the coefficient of L 

But since P + Qi = hj hypothesis, it follows from § 281, 
that both P = and Q = 0. We wish to prove that a — bi is 
a root of f(x) = ; i.e. f{a — bi) = 0. To prove this we 
merely have to notice that f(a — bi) may be obtained from 
the expression for /(a 4- bi) by replacing i by — i. Therefore 

f{a-bi)=P- Qi, 

where P and Q represent the same quantities as before. But 
we have just shown that P = and Q = 0. Therefore 
/(a — bi) = or a — 6i is a root oif(x) = 0. 

EXERCISES 

1. Solve X* + 4 a;3 + 5 a:2 + 2 X - 2 = 0, one root being - 1 + i. 

2. Solve a!:* + 4a:S + 6a;2-f4x + 5=:0, one root being i. 

3. Solve a;4 - 2 x3 + 5 a;2 — 2 a: + 4 = 0, one root being 1 — i Vs". 

4. If a+ Vb (a and b rational but V6 irrational) is a root off(x) = 
with rational coefficients, a — V6 is also a root. 

[Hint : Show that /(a + y/b) reduces to the form P + Qy/b where P 
and Q contain only integral powers of b and Q is the coefficient of Vb. 
Since P + QVb = 0, P = and Q = 0. Why ?] 

5. Solve 2 x* - 3 a;3 - 16 x2 - 3 X -f 2 = 0, one root being 2 + VS. 

6. Form an equation with rational coefficients, of which two of the 
roots are i and 1 + \/2. 

7. Solve the equation x^ - (4 + V3)x'^ + (5 + 4 V3)x - 5V3 = 0, if 
one root is 2 — i. 

8. Solve the equation x^ — (5 + i) x^ + (9 + 4i)x — 5 — 5 1 = if one 
root is 1 + i. Is 1 -- I a root ? 



464 MATHEMATICAL ANALYSIS [XIX, § 299 

299. To Multiply the Roots of an Equation. To trans- 
form a given equation f(x)= into another whose roots are those 
of /(a;) = each multiplied by some constant 7c, multiply the 
second term of f{x) by Jc, tJie third term by k^, and so on, taking 
account of the missing terms if there are any. 

The required, equation is f(y/k) = 0. For, if f(x) vanishes 
when X = a,f(y/k) will vanish when y = ka. Hence, if the given 
equation is a„a;" -f a^_^x''~^ + ... -f ao = 0, the required equation is 

""(!)" +""-■©"""'■■■ +""='' 

which on multiplication by k"" becomes 

ttn?/'* H- kan-if"-^ + k'^an-2y''-^ + • • • + k^ao = 0. 

If A: = — 1, we have, the roots of f(^x) = are equal respec- 
tively to those off{x) = with their signs changed. 

Example 1. Transform x^— ^x"^ -{- 5 = into an equation whose 
roots are twice those of the given equation. The desired equation is 
x^ - 4(2)a;2 + 6(2)3 = 0, or x^ - 8 x2 + 40 = 0. 

Example 2. Transform x'' — Sx^-\-4iX^-2x-\-l = into an equa- 
tion whose roots are those of this equation with their signs changed. 
The result is (-a:)7-3(-x)5 + 4(-x)4-2(-x) + 1 = 0, or 
a;7 _ 3 x5- 4 x* - 2 x - 1 = 0. 

EXERCISES 

Obtain equations whose roots are equal to the roots of the following 
equations multiplied by the numbers opposite. 

1. x6 - 2 x^ + x + 1 = 0. (2) 3. x4 - x2 + X + 1 = 0. (-3) 

2. x7-6x8+2x- 1 = 0. (-2) 4. x^ + x* -x^ + x- 1 = 0. (2) 

Obtain equations whose roots are equal to the roots of the following 
equations with their signs changed. 

5. x7 - 6 x« + 2 X* - X + 1 = 0. 7. x7 - x« + x6 - X* - 2 = 0. 

6. xi6 _ 1 = 0. 8. 1 - X - xa - x8 - X* - x6 = 0. 



XIX, § 300] GENERAL POLYNOMIAL FUNCTION 465 

300. Variations in Sign. A variation of sign or change of 
sign is said to occur in f(x) whenever a term follows one oi 
opposite sign. Thus the equation o^ — 3 a;^ 4-7=0 has two 
variations of sign. 

If f(x) has real coefficients and is exactly divisible by x — 7c, 
where k is positive, then the number of variations of sign in the 
quotient Q{x) is at least one less than the number of variations of 
sign infix). 

Before proving this statement let us consider the process of 

dividing /(a;) = x^-{- x^ — 3 x^— 2 x^ — x^ + b x—lhj x — 1 and 

/(x) = a;^ — a^-f- 4 a;2— 13 cc -h 2 by a; — 2, making use of synthetic 

division. 

• 1 1 _3 _2 -1 5 -1 g. 

1 2-1-3-4 1 

Q(a;) = l 2-1-3-4 1 

1 _1 4 -13 2 |2 
2 2 12 -2 



Q(a!)=l 16-1 

It will be noted in these examples that Q(x) has no varia- 
tions except such as occur in the corresponding or earlier 
terms of f{x) and that since f(x) is exactly divisible by the 
given divisor, the sign of the last term of Q{x) is opposite to 
that in f{x). Let us now prove the statement in general. 

Proof : From the nature of synthetic division it follows that 
the coefficients in Q(x) must be positive at least until the first 
negative coefficient of f{x) is reached. Then, or perhaps not 
until later, does a coefficient of Q{x) become negative or zero, 
and then they continue negative at least until a positive 
coefficient in f{x) is reached. Therefore Q{x) has no variations 
except such as occur in the corresponding or earlier terms of 
f(x). But by hypothesis f(x) is exactly divisible by a;— A; and 
2h 



466 MATHEMATICAL ANALYSIS [XIX, § 300 

hence the sign of the last term in Q(x) must be opposite to 
that in f(x). Therefore the number of variations of sign in 
Q(x) must be at least one less than the number of variations 
of sign in /(a;). 

301. Descartes's Rule of Signs. Theequationf{x)=0 with 
real coefficients can have no more positive roots than there are 
variations of sign in f(x) and can have no more negative roots 
than there are variations of sign inf(— x). 

Proof: Let r^, ^2, — , ^^(i? ^ ^) denote the positive roots of 
f(x) — 0. If we divide f{x) by a; — ri, the quotient by ic — rj, 
and so on until the final quotient Q{x) is obtained, then we 
know from the last theorem that Q(x) contains at least p fewer 
variations of sign than f{x). But the least number of varia- 
tions of sign that Q{x) can have is zero. Therefore f{x) must 
have at least p variations, i.e. at least as many variations as 
f{x) =0 has positive roots. 

Second, by § 299, we know that the negative roots of /(a;)=0 
are the positive roots of /(— a;) = and, hence, by the first 
part of this proof, we know that their number cannot exceed 
the number of variations of sign in/(— «). 

It is important to notice that Descartes's rule of signs does not tell us 
how many positive and how many negative roots an equation has. It 
merely tells us that an equation cannot have more than a certain number 
of positive roots, and cannot have more than a certain number of nega- 
tive roots. 

Example. What conclusions regarding the roots of the equation 
/gT _ 4 x5 + 3 a;2 — 2 = can be drawn from Descartes's rule ? 

The signs of /(a;) are -f 1 — , i.e. there are 3 variations and hence 

the equation has no more than 3 positive roots. 

The signs of /( — ic) are h + — , i.e. there are two variations and 

hence the equation has no more than 2 negative roots. 

But the equation is of degree 7 and has 7 roots. Therefore the equa- 
tion has at least two imaginary roots. Can there be more than two 
imaginary roots ? 



XIX, § 303J GENERAL POLYNOMIAL FUNCTION 467 

EXERCISES 

What conclusions regarding the roots of the following equations can 
be drawn from Descartes' s rule ? 

1. x^ - 2 x6 + X* - 1 = 0. 4. x^ - 2 X* + x^ - x^ - X + 1=0. 

2. x^ + X* — x^ + 1 = 0. 5. x** - 1 = 0. (w odd) 

3. x23 - 34 xi2 + X — 45 = 0. 6. x** - 1 = 0. (n even) 

7. Show that the equation x^ — 5x2 — x + 10 = has at least two 
imaginary roots. How many may it have ? 

8. Show that the equation x^ + x^-i-x — 1=0 has two and only two 
imaginary roots. 

9. Show that the equation x^ + 4 x' + 2 x — 10 = has six and only 
six imaginary roots. 

10. Can you tell the nature of the roots of the equation x* + ix^ — 3 ix 
+ 4 = 0? 

302. Equations in ^-form. If each term of the equation 

f{x) = a„a;" + a,,_,x^-' + ... + a^ = 

is divided by a„ (by hypothesis an ^ 0), we obtain the equation 

x^ + p^x''-'^ + PiX""-^ + ... + jp^ = 0, 

in which the leading coefiicient is unity and p^ = -^^, etc. An 

equation in this form is said to be in the p-form. For many 
purposes this is the most convenient form. 

303. Rational Roots. A rational root (^0) of the equation 
f{x) = when the equation is in the p-form with integral coeffi- 
cients is an integer and an exact divisor of the constant term. 

Proof. Suppose that the equation f(x)= has a root a/b 
where a/b {b > 1) is a rational fraction in its lowest terms. 
Then we have 

(11) (tf + p/tT' + ■■■+ pUt) +P, = 0. 



I, - " \b 



468 MATHEMATICAL ANALYSIS [XIX, § 303 

Multiplying both, members of (11) by 6"~^ we have 



or 

(12) T^~ (-^1^""' + ihci^'-^h + - + Pnh^"), 

The right-hand member of (12) consists of terms each of 
which is an integer. The left-hand member of (12) is a 
fraction in its lowest terms. Therefore the assumption 
that the fraction a/h is a root of f{x) = leads to an 
absurdity. 

Now suppose r (:^ 0) is an integral root. Then 

rn -hpir"-! -f p^r^-^ -+- ••• -h i?« = 0. 

If we transpose the constant term pn and divide by r, we 
obtain 

(13) r^-i^p^r^-2+ ... -|.p^_^ = _-P2. 

r 

Now each term of the left-hand member of (13) is an integer ; 
hence pn/r must be an integer, i.e. pn must be exactly divisible 
by r. 

304. To Find the Rational Roots of an Equation with Ra- 
tional CoeiKcients. If the equation is not in the p-form with 
integral coefficients, reduce it to that form and then make use 
of the results in § 303. The following examples will explain 
the methods. 

Example 1. Solve the equation x^ + S x^ — 4 x ^ 12 = 0. 

By Descartes's rule of signs we know that the equation has no more 
than one positive root and no more than two negative roots. From the 
last article we know that if the equation has rational roots they are 
factors of 12. Thus we need only try 1, — l, 2, — 2, 3, - 3, 4, — 4, 6, 
- 6, 12, -12. 



XIX, § 304] GENERAL POLYNOMIAL FUNCTION 469 

By synthetic division we have 

1 3-4-12 12 

2 10 12 I 
16 6 

The depressed equation * ia x^ -\- ^ x + 6 = (x -^ S)(x + 2)= 0. There- 
fore the roots of the original equation are 2, — 3, — 2. 

Example 2. Solve the equation 2a:8_^a;2 + 2x + l = 0. 
Writing the equation in the |)-f orm we have 

X3 + I X2 + X + ^ = 0. 

If we multiply the roots of this equation by k, we obtain 

x^-\--kx^-\-k'^x + — =0. 
^ z 

If we choose k equal to 2, this equation becomes 
(14) x^-\-x^-h4:X + 4 = 0, 

an equation whose roots are twice those of the original equation. 

By Descartes's rule of signs equation (14) has no positive roots. Any 
rational roots are then negative, and are factors of 4, i.e. — 1, — 2, — 4. 
By synthetic division 

1 14 4 |-1 

-10-4 
10 4 
The depressed equation is a;^ + 4 _ q. Therefore the roots of (14) are 
— 1, 2 1, — 2 I and the roots of the given equation are — ^, i, — i. 

EXERCISES 
Solve each of the following equations. 

1. ic« + 5 x2 -f- 15 X + 18 = 0. 4. 6 a;3 + 7 x2 _ 9 x + 2 = 0. 

2. xs + x2 + X + 1 = 0. 5. 6 x3 - 2 ic2 + 3 X - 1 = 0. 

3. x=i + x2-4x-4 = 0. 6. 2x4 + 3x3-10x2-12x+8=0. 

Find the rational roots of each of the following equations. 

7. X* - 3 x2 - 4 = 0. 10. 2 X* - x8 - 5 x2 + 7 X - 6 = 0. 

8. x6 - 32 = 0. 11. 2 X* + 2 x3 - x2 + 1 = 0. 

9. X* + x8 + x2 + X + 1 = 0. 12. 4 x* - 23 x2 - 15 X + 9 = 0. 

* If r is a root of a given equation /(.x*) = and f(x) = (x — r)Q(x), then the 
equation Q{x)= is called the depressed equation. 



470 



MATHEMATICAL ANALYSIS [XIX, § 305 



306. The Solution of an Equation with Numerical Coeffi- 
cients. The preceding articles furnish a number of methods 
for attacking the problem of finding the roots of an algebraic 
equation/ (a;) = with given numerical coefficients. 

(1) We may examine the equation for multiple roots (§ 297). 

(2) If the equation f{x) = has rational coefficients, we can 
find all the rational roots by a finite number of trials. 

(3) When any root a has been found, we may divide f(x) by 
X — a and thus make the finding of the remaining roots depend 
on an equation of lower degree (the depressed equation). 

306. Irrational Roots. Graphical Approximation. In order 
to compute approximately any one of the real irrational roots 
of an equation/ (a;) = whose coefficients are real numbers, we 
require first a rough approximation to the root which is to be 
computed. The graph of y =f(x) is a powerful tool for this 
purpose. An example will make the method clear. 



Example. Locate approximately the real roots of the equation 

f(x) = x6 - ISx'-^ + 2 a; + 5 = 0. 
A table of corresponding values of x and 
f(z) is as follows. 



I^' 


-4^ 




7ko_ 


::::^:i::i-:: 


:i:::|?:i::d=: 


::::4?:i::: i: 


-^ f- 




1)^^ uy 


:E/:E:::E:E:E: 



X -2 - 1 


1 


2 3 


/(a;) -83-11 6 


-5 


- 11 137 



Fig. 262 



Figure 252 exhibits a rough graph of this 
function constructed from this table. We 
conclude that a root of the equation lies 
between — 1 and 0, another between and 1 , 
and a third between 2 and 3. 

Moreover Descartes 's rule tells us that this 
equation can have no more than two positive 
roots and no more than one negative root, 
since there are only two changes of sign in 
f(x) and only one in/(— x). 

We have therefore located all the real roots 
of this equation. 



XIX, § 307] GENERAL POLYNOMIAL FUNCTION 471 



A more accurate construction in the neighborhood of one of these points 
enables us to get a better approximation. For example, the values x = 2.2 
and 2.3 give us respectively y = — 1.97 and 6.21. By drawing a smooth 














~ 








-T-- 












































4* -. 






















^^ 






















" ^: : 






















^^ 


-rr 


















,t 








2-0 






-2- Ir- 








2-e^' 
,../ 




' 








































^ i* 






















^^' 




















,'' 






















** 






































, * 






















*' 
















— 




*^^ - 


- __ 


- 















"- 








- 


-■ 











Fig. 253 

curve through the three points corresponding to x = 2, 2.2, 2.3 (plotted 
on a large scale, Fig. 253) we may estimate the root of f{%) = to be ap- 
proximately 2.23. 

307. Newton's Method of Approximation. Having found a 
first approximation to a root of an equation /(«)= 0, we may 
secure a better approximation by a 
method first suggested by Sir Isaac 
Newton (1642-1727). In Fig. 254 let 
GC represent the graph of y—f{x) in 
the neighborhood of a root a? = a of the 
equation. Let OM^ — x^ represent the 
approximation to the root found ; let 
M^Px = ^1= /(xi). Let the tangent to the 
graph at Pi(iCi, y-^ cut the a;-axis in T. 
The abscissa OT will then, in general, be a much closer 
approximation to the desired root. The equation of the 
tangent at Pj is 
(15) 2^-/(^i) = /'(^0(^-a^i). 

Placing 2/ = and solving for x we have 




Fig, 254 



(16) 



052= OT=Xy 



/fe) 



472 MATHEMATICAL ANALYSIS [XIX, § 307 

where x^ denotes our second approximation. We have then 

(17) ^2 = JCi + ^1, 
where the correction hi is given by 

(18) K = -f^. 

Example. Find by Newton's method a better approximation to the 
root X = 2.23 of the equation x^ — 13 x''^ + 2 ic4- 5 = discussed in § 306, 

f(x) = ic5-13x2 + 2a; + 5. 

/'(x) = 5a:*-26x + 2. 

/(xi)=/(2.23)=-0.039.» 

/'(xi)=/'(2.23)=67.67.» 
Hence we have 

^~ /'(a^i) 67.67 
whence X2 = 2.23057. 

308. The Accuracy of Newton's Method. A question that nat- 
urally arises is : How accurate is this root, i.e. to how many decimal 
places is it correct ? Taylor's theorem gives us information on this point. 
We have ,^ 

(19) /(xi + ;iO = /W + /'(^i)'^i+'^-^^^i'+ -. 

If our first approximation to the root is x = Xi and hi is the correction, f 

Newton's method gives to ^i a value which makes the sum of the first two 

terms of Taylor's expression vanish. Since h\ is very small, the terms 

beyond the third (involving h\^ and higher powers of ^i) are insignificant 

fii (xA 
compared with the term , . h^. Hence for our purpose we may write 

2i I 

(20) /(a:i + Ai) = i/"(a;i)A;2. 

In the example considered above we have 

/"(x)= 20x3-26. /"(a^i) =/"(2.23)= 195.8. 

hi^ = (0.00057)2 = 0.00000032. 
Hence we have 

\f"{x{)hi^ =/(xi + h{) = 0.0000313. 

♦ Use synthetic division to get these values. 
t In the example just cousidered ^i = 0.00057. 



XIX, § 309] GENERAL POLYNOMIAL FUNCTION 473 

Moreover ^,^^^ ^ ^^^ ^^,^^^^ +r{x{)h, + ..-, 

and in this example /'(aii + hi) = 67.67 + 0.11 = 67.78 approximately. It 
follows that the new correction is about 

;i2 = ^/(?l+M<_ 0.000001. 
f'{xi + hi)^ 

Therefore we may conclude that x = 2.23057 is the root sought, to five 
decimal places. 

EXERCISES 

Find to three places of decimals the irrational roots of the following 
equations. 

1. a;3 + 3x + 20=0. 2. a;^ + 2x2 - a; + 3 = 0. 3. a^ + a;-l = 0. 
4. a:8 + 4 a;2 _ 6 = 0. 5. x^ + Sx"^ - Sx-1 = 0. 

6. If X is the cosine of an angle and y is the cosine of one third of the 
angle, then 4 y^ = 3 ?/ + x. Find the value of cosine of 20° to three places 
of decimals. 

7. An open box is to be made from a rectangular piece of tin 9 x 10 
inches, by cutting out equal squares from the corners and turning up the 
sides. How large should these squares be so that the box shall contain 59 
cu. in,? 

8. Find the cube root of 12 ; 45 ; - 37. 

309. The Relation between Roots and Coefficients. If 
fit ^2) •••? ^„ are the roots of the equation x"" + picc""^ 4- P2^"~2 _|_ 
— hPn = 0, then 

x^ -{- PiX""-^ -\- PiX""-^ + ... -\-p„=(x-r{){x-r2)-{x-r,;). 

If we carry out the indicated multiplication in the right-hand 
member and equate the coefficients of like powers of x, we ' 
have 

(21) Pi = -ri-r2 r,, 

(22) JP2 = nrz + nra + ... + nr^^ + r^n + - + »*n-iV 

(23) Ps = - ^1^2?*3 - r^r^U ^n-2^n-l^n. 

(24) P«=(-l)Vir,...r,. 



474 MATHEMATICAL ANALYSIS [XIX, § 309 

That is, 

— Pi= the sum of the roots. 

JP2 = the sum of the products of the roots taken two at a time. 

— Pg = the sum of the products of the roots taken three at a time. 

(— lyPn = the product of all the roots. 

We have at once the following corollaries : 

1. To transform an equation into another whose roots are those of the 
original equation each midtiplied by m, multiply p\ by m, p2 by m^, p^ by 
m^, and so on (§ 299). 

2. To transform an equation into another whose roots are equal to 
those of the original equation with their signs changed, change the signs 
of the alternate terms, beginning with the second. 

Example 1. Solve the equation 2y:^— x'^ — 9>x + ^ = Q given that 
two of the roots are equal in absolute value but opposite in sign. 

Let the roots be r, — r, and s. 
Then r — r + s = \, 

rs — rs — r^ = — 4:, 
-r2s=-2. 
Therefore s = ^ and r = 2 or — 2, i.e. the roots are ^, 2, — 2. 

EXERCISES 

1. Solve a* + x* — 4 5c — 4 = 0, given that the sum of two of the roots 
is zero. 

2. Solve x^ — 6 a;3 — 9 x2 + 54 x = 0, given that the roots are in arith- 
metic progression. 

3. Solve X*- 16 x3+ 86 x^- 176 x + 105 = 0, given that the sum of two 
roots is 4. Ans. 1, 3, 5, 7. 

4. Solve 4 x8 — 20 x2 — 23 X — 6 = 0, two of the roots being equal. 

5. If n, ro, rs are the roots of x^ — 5 «'-* + 4 x — 3 = 0, find the value 
of each of the following expressions : 

(a) ri2 + rz^ + n^. 

(6) ri^ -\- r2^ + ra,^ 

(c) riV + ri^rs^ + r^^rs^. 

(d) ri^ri + r-i^r^ + r2'Vi + rz'^rs + rs^i + r^^r^. 



CHAPTER XX 
DETERMINANTS 

310. Determinants of the Second Order. Expressions of 
the form aib2 — a^bi , where a^, tt2, 61 , 62 ^'^^ ^^J numbers, arise 
often in mathematical analysis. Thus the area of a triangle 
with one vertex at the origin and the other two vertices at the 
points (ai , bi), (a2 , 62)? is equal to ^{aib^ — a26i) (§ 195). Again, 
the solution of a pair of simultaneous linear equations in two 
unknowns (§ 69) can be written as two fractions whose numer- 
ators and denominators are all of this form. (§ 311.) 

The expression aib^ — ^2^1 ^^J ^^ written in the form 

I ai bA 

and is then called a determinant of the second order. Such a 
determinant contains two rows and two columns. The numbers 
«i > «2 J ^1 J ^2 J are called the elements of the determihant. The 
two elements ai, 62 form the so-called principal diagonal. 

To evaluate a determinant of the second order, i.e. to find 
what number it represents, one merely has to subtract from 
the product of the terms in the principal diagonal the product 
of the other two terms. Thus we may write 



tti bi 

0"2. ^2 



= aib^ — a^bi ; 



4 7 

3 -6 



= (4)(-6)-(3)(7)=-45. 



It is important to notice that each term of the expansion 
contains one and only one element from each row and one and 
only one element from each column. 

475 



476 



MATHEMATICAL ANALYSIS [XX, § 311 



311. Simultaneous Equations in Two Unknowns. Let the 

equations be 



(2) 



1 OgflJ + 62S/ = C2. 



If we solve these equations by the usual method of elimination 
(§ 69), we obtain 



(3) 



C162 ~ ^2^1 ^l^ — ^2^ 

"" aib2 — a^hi ' "" 0162 — a2&i ' 



provided a^h^—a^hi^O. We at once recognize the fact that 
these results may be written in the form 



(4) 



x = 



Ci 61 




ai Ci 


C2 62 




02 C2 


«! h 


'} y — 


«! &1 


^2 &2 




02 &2 



provided 0162 — c^h =^ 0. The following points should be noted 
in the above solution. 

(1) The determinants in the denominators are identical and 
are formed from the coefficients of x and y in the original 
equations. . 

(2) Each determinant in the numerator is formed from 
the determinant in the denominator by replacing by the 
constant terms the coefficients of the unknown whose value 
is sought. 

Example. Solve by determinants the simultaneous equations 
(2x-y = l, 
[Sz + 2y = S. 



Solution : 



1 


-1 




2 


1 


3 


2 


6 
= 7' y=- 


3 


3 


2 


-1 


2 


-1 


8 


2 




3 


2 



XX, § 311] 



DETERMINANTS 



477 



EXERCISES 

Evaluate each of the following determinants 

^ 14 61 ^ I — sin a — cos a I 

1 • 3. 

1 3 1 1 I cos a sin a I 

12 a 61 Itan^ sec^l 

c — d\ ' sec ^ tan d 



I sin e cos 6 
sin a cos a 



2. 



6. Show that the normal form of the equation of a straight line 
(§ 205), may be written in the form 

.* " Up. 

— sm a cos a \ 
Solve by the use of determinants the following pairs of equations : 



f2x + y = 3, g 

\ 6x^ y = 4. 



Ax- Sy_ 



= 2, 



¥— • 



9. 



x + y = l, 
3 *^ 3 



10. 



11. 



ic sin ^ 4- y cos ^ = sin 0^ 
xcosd ■{■yemd = cos 6. 
X -\- y tan ^ = sec^ 6, 
X sec2 d +yctne = sec* ^ + 1- 



^ns. 1, tan e. 



Prove the following identities and state in words what they show. 



12. 



13. 



14. 



15. 



16. 



ai &i 




Ol 


02 


0,2 62 




61 


62 


dl Oi 

02 O2 


= 


0. 





ai 61 
a2 62 
mai ?)i 
maa 62 
(ai + 5i) 61 

(a2 + &2) &2 



= r» 



61 ai 

62 02 
«i &i 

02 62 

aa 62 



* For example, Ex. 12 shows that in a second-order determinant if the cor- 
responding rows and columns are interchanged, the value of the determinant 
is not changed. 



478 



MATHEMATICAL ANALYSIS [XX, § 312 



312. Determinants of the Third Order. To the square 
array 

bi Ci 



(5) 



O2 C2 
^3 Cs 



we assign the value 

(6) aidgCg + azhci + ag^iCz — 016302 — aa^iCg — a^bzCi 

and the name determinant of the third order. 

The expression (6) is known as the development or expan- 
sion of the determinant, the numbers ai, hi, etc., as the elements, 
and the elements Oi, 62, Cg as the principal diagonal. 

It is important to notice that in the development (6) each 
term consists of the product of three elements, one and only- 
one from each row and one and only one from each column. 

An easy way of obtaining the expansion (6) of the deter- 
minant (5) is as follows : 

Form the product of each element of the first column by the 
second-order determinant formed by suppressing both the row 
and column to which the element belongs. Change the sign of 
the product which contains the element in the first column and 
the second row and take the algebraic sum of the three products. 



Example 1. 



Example 2. 



Example 3. 



a\ 61 ci 
















62 Co 


hx Ci 




&1 Ci 


ai 62 C2 


= a\ 




-a'i 


+ as 








hz ca 


hz cs 




62 C2 


az 63 C3 













ai&2C3 — ai&3C2 — «2^iC3 + «2&3Ci + a^biCi — azbiCi. 



2 3 
-6 4 

4 -1 



3 1 
-1 
6 



+ 6 



il + ^U 



= 2(4 + 7)-H 6(3 + 2) + 4(21 - 8)= 99. 



= 3 



= 3(_2-20): 



m. 



XX, § 312J 



DETERMINANTS 



479 



EXERCISES 
Evaluate each of the following determinants. 



1. 



2 


1 


3 


1 


1 


-1 


1 


1 


2 



2. 



-1 





1 




5 


-2 


-2 


2 





. 3. 


6 


-3 


3 


5 


1 




7 


-4 



41 



la 6 c 
6 <r c 
c a 6 



6. In § 196 it was shown that the area of the triangle whose vertices 
are Pi(a:i, yi), P2(2C2, 2/2)1 ^3(^:3, tjs) is 

^[Xi?/2 - X2?/l + 3^22/3 - X3?/2 + Xa^l - Xi^s]- 

Prove that the area of this triangle is 

xi yi 1 
I X2 y2 I ■ 

X3 VZ 1 

6. Using the result of Ex. 5, find the area of the triangle whose 
vertices are 

(a) (2, 1), (3, !),(-!, 7); 
(6) (3,2), (3, 6), (-1,-4); 
(c) (0, a),.(0, -a), (&, 0). 

7. Prove that the three points Pi(xi, ?/i), P2(a;2, 2/2), P3(a53» ys) are 
coUinear if, and only if. 



a^i yi 


1 


X2 y2 


1 


X3 yz 


1 



8. By means of determinants show that the three points (a, & + c), 
(&, c -\- a), (c^ a + b) are collinear. 

9. By use of determinants determine whether the three points (0, 0), 
(1, 1), (5, 6) are collinear. 

10. Prove that the equation of the straight line through the points 
Pi(a:i, t/i), P2(X2, 2/2), is 

= 0. 



X 


y 


1 


Xi 


yi 


1 


X2 


2/2 


1 



11. By determinants find the equation of the straight line through each 
of the following pairs of points! 

(a) (2, 1), (3, 7) ; (6) (6, 1), (2, - 1); (c) (7, 1), (9, 1). 

12. Find by the use of determinants whether the three lines 3 a;— y — 7 
= 0, 2x4-j/ + 2=:0, a: — y = are concurrent or not. 



480 



MATHEMATICAL ANALYSIS [XX, § 313 



equations be 
(7) 



313. Solution of Three Simultaneous Equations. Let tha 

aiOJ 4- 6i2/ + CiZ = dj, 

If we solve these three simultaneous equations by the usual 
method of elimination, we obtain, 

dibzC^ + dzbsCi + dsbiCz — dzb2Ci — dih^^Cz — d^bic-^ 



(8) 



~~ ciiboCs + ota^gCi + a^biCz — a^bzCi — aib^Cz — azbic^ 
_ aid2Cs + a2C?3Ci + a^diC2 — a^doC} — ai(?3C2 — aodiC^ 



z = 



«]&2C?3 + «2^3C?1 + «3^1<^2 — «3^2f'l 



«i&2C3 + (hbsCi -\- (136102 — 0362^1 



O163CZ2 — OL^id^ 



^1 


&1 


Ol 


C?2 


62 


Ci 


C^3 


&3 


C3 


«1 


bi 


Cl 


tti 


b. 


C2 


<h 


h 


C3 





cti 


^1 


Cl 




<H 


(i2 


C2 


V —r. 


as 


d. 


C3 


if 


«! 


bi 


Cl 




a2 


&2 


C2 




^3 


^^3 


C3 





ai 


&i 


C?i 




ag 


&2 


d, 




^3 


63 


C?3 




ai 


^^1 


Cl 




ag 


60 


C2 




03 


&3 


C3 



provided the denominator of each fraction is not zero. These 
results may be written in the form 



(9) 



Each denominator is the same determinant, which is called 
the determinant of the system. It is made up of the coefficients 
of X, tfj z. Each determinant in the numerator is formed from 
the determinant in the denominator by replacing the coefficients 
of the unknown whose value is sought by the constant terms. 
Compare this rule with that given in § 311. 

ExAMPLB. Solve the following equations by determinants : 

f6x-22=-2, 
-3y-4;2 = 7, 
2x-6j/ = -10. 



XX, § 313] 



DETERMINANTS 



481 



Solution. 

-2 -21 

7 -3 -4 

-19-5 ol 



u 


-2 


-3 


-4 


-5 






224 



112 



= -2;2/ = 



5 


-2 


-2 







7 


-4 




2 


-19 





-336 


5 





-2 


- 112 





-3 


-4 




2 


-6 








5 





-2 





-3 


7 


2 


-5 


-19 


5 





-2 





-3 


-4 


2 


— 5 






448 
-112 






3. 



EXERCISES 
Expand each of the following determinants : 



4. 



Solve by determinants each of the following sets of equations ; 



2 3 

4 -1 

-1 4 


3 

2 

1 


1 1 1 
a h c 
a2 62 c2 





-7 1 
2 -2 
4 2 


2 

► -6 

4 


a h g 
h b f 
g f c 





a; 


z 


z 


a; 


y 


y 


y 


z 


X 



[4aj4-5y + 22r = 20, 
6. j 3 X - 3 ?/ + 5 5! = 12, 

[5x + 22/-40=-3. 

Ans. (1, 2, 3). 

[x-^y ■\-z=\, 
8. ax -f- 6y + cs = d, 

[ cC^x + 6'^?/ 4- c-z = d^. 

10. Solve the equation 



7. 



3x + y-z = 3, 
x + y + z =7, 
2x + 4y -\-z=12. 

ax + y — z = a'^ + a — l, 
— x + ay-\-z = a^— a-\-l, 
x — y + az = a. 

Ans. (a, a, 1), 



1 

-1 

5 



0. 



11. Solve for x and y the simultaneous equations 



2i 



x+1 
3 

y 



2 1 

X -1 

2 1 



= 0, 



X 

-2 

y 



0. 



482 



MATHEMATICAL ANALYSIS [XX, § 313 



12. Evaluate the determinant 

sin a cos /S 1 

cos a sin /3 1 

1 1 1 

Prove the following identities and express in words what they prove. 
See Ex. 12-16, pp. 477. 



13. 



15. 



17. 



«! &i Ci 

a^ hi C2 

az h C3 

«! tti &i 

Gi a^ bi 

as «3 &3 



ai 


tti «3 




ai 61 Cl 




bi 


bz 62 


. 14. 


a2 62 C2 


= - 


Cl 


C2 C3 




^3 bs C3 





ai 


bi 


Cl 


as 


bz 


C3 


a2 


62 


C2 



0. 



16. 



max 61 Cl 
ma2 62 C2 
maz bz cz 



ax 


61 


Cl 


as 


&2 


02 


as 


68 


cs 



(ai + 61) 61 Cl 
(«2 + 62) 62 C2 
(as + bz) bz C3 



ai 


bx Cl 


a2 


62 C2 


az 


bz cz 



314. Inversions. Let us consider the permutations of a set of ob- 
jects, such as letters or numbers, and let us fix a certain particular order 
of the objects which we shall designate as the normal order. An inversion 
is said to occur in any permutation when an object is followed by one 
which in the normal order precedes it. Thus if abed is the normal order, 
then there are two inversions in bade. If 1234 is the normal order, then 
there are three inversions in 1432. 

Theorem. If in a given permutation, two objects are interchanged, 
the number of inversions with respect to the normal order is increased or 
decreased by an odd number. 

Let us consider the permutations Xrs Y and Xsr Y, where X and ;i" 
denote the groups of objects which precede and follow the interchanged 
objects r and s. Any inversion in Xand Fand any inversi(m due to the 
fact that X, r, s precede Y are common to Xrs Y and Xsr Y. Therefore, the 
number of inversions in Xrs Y is equal to the number in Xsr Y increased 
or decreased by 1 (according as rs is or is not in the normal order). 

Now let us consider two objects such as r and s separated by i objects. 
If the objects r and s are interchanged, the number of inversions is still 
changed by an odd number. For, by f + 1 interchanges of adjacent pairs 
the object r can be brought into the position immediately following s, and 
by i further interchanges of adjacent pairs, s may be brought to occupy 



XX, § 316] 



DETERMINANTS 



483 



the position formerly held by r. Each of these (i -f 1) + i = 2 i 4- 1 
interchanges of adjacent pairs has increased or decreased the number of 
inversions by 1. Hence the net result of these 2^ + 1 interchanges has 
increased or decreased the number of inversions by an odd number. 

315. Determinants of the nth Order. The square array 



(10) 



ai 6i 
a2 62 



an K 

of n^ elements, such as we have considered for the cases n = 2, w = 3, is 
called a determinant of the nth order and will be denoted by the Greek letter 
A. This determinant will he understood to stand for the algebraic sum of 
all the different products of n factors each that can be formed by taking 
one and only one element from each row and one and only one element 
from each column, and giving to each such product a positive or negative 
sign according as the number of inversions of the subscripts {normal 
order 1, 2, •••, n) is even or odd, when the letters have the normal order 
ab -'-q. 

It should be noted that from the remarks in § 314 it follows that if we 
arrange the elements in any product so that the subscripts are in normal 
order, we can determine the sign of each term, by making it positive or 
negative according as the number of inversions of the letters is even or odd. 

316. Properties of Determinants. Theorem 1. The expan- 
sion of a determinant of order n contains n ! terms. 

Proof. There are as many terms in the expansion of a determinant 
of the nth order as there are pennutations of the subscripts 1, 2, 3,---, n. 
But this number is n ! (§ 269) . 

Theorem 2. If each element of any row or column is multiplied by any 
constant m, the value of the determinant is multiplied by m. 

Proof. Since by the definition of a determinant, each term of the ex- 
pansion must contain one and only one element from each row and each 
.column, the factor m will appear once and only once in each term of the 
expansion. If m is factored out of this expansion, the remaining factor 
is the expansion of the original determinant. 



484 MATHEMATICAL ANALYSIS [XX, § 316 

Illustration. 

inaib2C3-\-ma2b3Ci+mazbiC2—maibzC2 — mazbiCz — mazbiPi 



moi 


bi 


Cl 




mctz 


b2 


C2 


= mt 


maz 


bz 


cz 


= m 



ai bi 


Cl 


02 bz 


C2 


az bz 


cz 



Theorem 3. The value of a determinant is not changed if roics and 
columns are interchanged, so that the first row becomes the first column, 
the second row the second column, and so on. 

This follows at once from the definition of the determinant and the 
paragraph immediately following it (§ 815) . 

Theorem 4. If two rows or two columns of a determinant are inter- 
changed, the sign of the determinant is changed. 

Illustration. See Ex. 14, p. 477, and Ex. 14, p. 482. 

Proof : Since by Theorem 3 rows and columns may be interchanged 
without affecting the value of the determinant, we need only consider the 
interchange of two rows. First, if two adjacent rows are interchanged, 
the order of the letters in the principal diagonal and in each term of the 
development is left unchanged. However two adjacent subscripts in each 
term of the expansion are interchanged, and hence the sign of every term 
is changed. Why ? 

Next consider the effect of interchanging two rows separated by k inter- 
mediate rows. By k interchanges of adjacent rows, the lower row can be 
brought just below the upper one. Now the upper row can be brought 
into the original position of the lower row Toy k -\- 1 further interchanges 
of adjacent rows. Therefore interchanging the two rows is equivalent to 
2k-\-l interchanges of adjacent rows. But 2 A; + 1 is an odd number and 
therefore this process changes the sign of the determinant. 

Theorem 5. If two rows or two columns of a determinant are identical, 
the value of the determinant is zero. 

Proof : Let A be the value of the determinant and let the two identi- 
cal rows or columns be interchanged. Then, by Theorem 4, the value of 
the resulting determinant is — A. But since the rows or columns which 
were interchanged were identical, the value of the determinant is left 
unchanged. That is to say, A = — A or 2 A = 0, or A = 0. 

Corollary. If all the elements in any row or column are the same 
multiples of the corresponding elements in any other row or column, then 
the value of the determinant is zero. 



XX, § 317] 



DETERMINANTS 



485 



317. Minors. If we suppress the row and the column in which any 
given element appears, the determinant formed by the remaining elements 
is called the minor of that element. 

Illustration. In the determinant 



the minor of az is 



ax 


6i 


Cl 


az 


&2 


C2 


«3 


&3 


cz 



and the minor of Cs is 



The minor of a\ is denoted by A\, of hj by ^y, etc. 



hi 


Cl 


bs 


Cs 


«i 


bi 


Cl2 


62 



EXERCISES 



1. Prove that 


2 2 3 
1 1 5 
4 4 9 


= 


0. 


2. 


Prove that 


3. Prove that 


4 5 6 
2 1 5 
1 5 3 


= 


4 2 1 

5 1 5 

6 5 3 


• 


4. Prove that 


3 4 5 
2 4 1 

8 4 5 


' = - 


4 3 5 
4 2 1 
4 8 5 


• 


6. Prove that 


4 1 
3 -1 

2 2 
1 3 

5 4 


6 5 3 
5 -1 -3 
2-3 6 
3 2 9 
-1 1 12 


= 0. 


6. Prove that 


26 9 
28 18 
30 3 


< 


-5 

10 
25 


= 30 


13 
14 
15 


3 -1 
6 2 
1 -5 



3 5 8 1 

1 2 5 =0. 

2 4 10 



7. How many inversions are there in the arrangement 4213765 if the 
normal order is 1234567 ? 

8. How many inversions are there in the arrangement 46321 if the 
normal order is 42316 ? 



486 



MATHEMATICAL ANALYSIS [XX, § 317 



9. Find the value of the minor of 5, of 6, of 7, for the determinant 
4 5 1 
3 6 2. 
2 7 8 

10. Write down the minor of as, of C2, of 64, for the determinant 

[1 &i ci di 
I2 62 C2 di 

64 C4 d^ 



11. Show that 
1 2 5-1 

3 3 6 2 

4 2 7 3 

5 15 4 




1 


6 4 3 




2 
5 


12 3 

5 7 6 


= - 


1 


4 3 2 





2 12 3 

15 4 3 

14 3 2 

5 5 7 6 



318. Additional Theorems. — The following theorems will be 
found useful in evaluating determinants. 

Theorem 6. Laplace's Expansion. If the product of each element 
in any row or column by its corresponding minor be given a positive or 
negative sign according as the sum of the number of the row and the num- 
ber of the column containing the element is even or odd, then the algebraic 
sum of these products is the value of the determinant. 

Proof : First, it is evident that in the development of the determinant, 
Ai is the coefficient of ai. For Ai is a determinant of order n — 1 in the 
elements a2, •••, a„, and its expansion contains a term for each permuta- 
tion of 2, 3, ••-, n. Moreover, the signs of the terms are correct ; for, the 
number of inversions is not changed by prefixing ai. 

Second, let us consider the element e situated in the zth row and the jth 
column. We can bring this element to the leading position, i.e. first row 
and first column, by i — 1 transpositions of rows and j — 1 transpositions 
of columns, i.e. hji-\-j — 2 transpositions in all. Therefore the sign of 
the determinant will have been changed i + j — 2 times. That is, if i + j 
is an even number, the sign of the determinant is left unchanged ; while if 
t+jis an odd number, the sign of the determinant is changed. Now 
that the element under consideration is in the leading position, we know 
from the first step that its coefficient is its minor. Since the relative posi- 
tions of the elements not in the ith row or the jth column are not effected 
by these transpositions, the minor of the element in its original position 
is the same as the minor of the element when it is in the leading position. 



XX, § 318] 



DETERMINANTS 



487 



Hence the coefficient of the element e, which is situated in the ith. row 
and the jth column, is (— ly+J E, where E is the minor of the element e. 

Corollary, If in the development of a determinant by minors with 
respect to a certain column (row) the elements of this column {row) are 
replaced by the corresponding elements of some other column {or row), 
the resulting expression vanishes. 



= aiAi — azAi + dsAs — a^A^i. 



We wish to show that, for example, 61^1 — 62^2 + 63^3 — b^Ai is 
zero. This expression is zero, for we have replaced the column of a's by 
the column of ?)'s and hence the determinant has two columns identical. 
The same proof applies to a determinant of order n. 

Theorem 7. If each of the elements of any row or column of a deter- 
minant consists of the sum of two numbers, the determinant may be 
expressed as the sum of two determinants. 

Proof : Let 



Illustration. 


ai 


bi 


Cl 


di 




a2 


62 


C2 


d2 




as 


63 


ca 


da 




a4 


64 


C4 


d. 



(ai + a'l) 
(a2 + a'2) 



{an + a'„) 6, 



Qn 



be the given determinant. Expanding in terms of the first column we 

have 

(ai + a'i)Ai - (aa + ^'2)^2 + (aa + a'3)^3 +••• + (- 1)'*-K«» + «'n)^n 

or [aiAi — a^Ai + azA^ + ••• + (— l)"-ia„A] 

+ [a'1^1 - a'2^2 + a'3^3 + ••• + (- l)"-ia'„J„], 



«1 


bi 


■ qi 




02 


62- 


■'; 


+ 


an 


bn- 


•Qn 





a'l 



Theorem 8. If to the elements in any row {or column) be added the 
corresponding elements of any other row {or column) each multiplied by 
a given number m, the value of the determinant is unchanged. 

The proof of this theorem follows easily from Theorems 7, 6, and 2. 



488 



MATHEMATICAL ANALYSIS [XX, § 319 



319. The Evaluation of Determinants. We are now in a posi- 
tion to expand a determinant of any order. The following examples will 
illustrate the methods employed. 

Example 1. Expand 



A = 



Multiply the first column by 
columns. It gives 

A = 



27 
28 
29 

1 and add it to the second and third 



25 


26 


26 


27 


27 


28 



25 


1 


2 


26 


1 


2 


27 


1 


2 



By the corollary of Theorem 5, the value of this determinant is 0. 

Example 2. Expand the determinant 

2-16 



A = 



1 

14 6 3 
4 2 7 4 
3 12 5 



We seek to transform this determinant in such a way as to make all 
the elements but one in some row or column 0. The second column 
looks most promising. We accordingly add 4 times the first row to the 
second row (this replaces the 4 in the second row by 0); we then add 2 
times the first row to the third row (Why?) ; and then add the first 
row to the fourth row (Why ?) These operations give 



A = 



2-1 5 1 

9 26 7 

8 17 6 

5 7 6 





9 26 7 




— 


8 17 6 
5 7 6 


— 



2 26 7 
2 17 6 
1 7 6 



The last determinant may be still further simplified as follows ; 



A = 



2 26 7 

2 17 6 

-17 6 

__ 140 191 
-"131 18|- 
= _ (162-31): 



40 19 

31 18 

-17 6 

9 1 

31 18 

131. 



* This determinant is obtained from the preceding by subtracting the 
elements of the last column from those of the first. 



XX, § 319] 



DETERMINANTS 



489 



EXERCISES 

Evaluate the following determinants. 



5. 



14 


13 


— 


121 




17 


16 


17 


. 


25 


24 


-18 




34 


23 


12 




23 


34 


21 


. 


14 


35 


26 




18 


26 


24 




29 


39 


49 


, 


37 


35 


11 




2 - 


-2 


1 1 




1 - 


-1 


4 2 




2 - 


-2 


1 -1 


• 





2 


1 -1 




3 


4 


-2 6 


4 - 


-3 


8 -4 


2 


8 


3 


1 





4 1 


23 


24 


25 26 


12 


13 


14 15 


32 


33 


34 35 


2 


2 


2 




2 



7. 



8. 



b c + d 
c b + d 
d 6 + c 

2 a a^ 

a + b ab 

2b 62 



abed 
6x00 
c y 
d z 



10. 



11. 



1 


1 


1 




a 


6 


c 


. 


a2 


62 


C2 




1 


1 


1 


1 


a 


6 


c 


d 


a2 


62 


C2 


(22 


a8 


68 


C« 


(28 



12. abed 
a 6 c (2 
a — 6 c 5 
a — 6 — c (2 

13. Prove that if a determinant v^hose elements are rational integral 
functions of some variable, as y, vanishes when y = 6, then y — b is a 
factor of the determinant. 

[Hint : Use the corollary of theorem 6.] 

14. Solve by factoring Examples 10, 11, 12. 

15. Factor into two factors 
a b c 
b c a 
cab 

16. Factor 



a 


a2 


be 


b 


62 


ca 


c 


C2 


ab 



490 



MATHEMATICAL ANALYSIS 



[XX, § 320 



320. Solution of a System of Linear Equations. Suppose we 
have n linear equations in n unknowns and we desire their solution. Let 
the equations be 



(11) 



aixi H- 6ia;2 + Cia-3 + 
a^Xi -f biXi + C2Xs + 



+ P\Xn = gi 



QnXl + \X2 + CnXs + ••• + PnXn= ^n 

Let A be the determinant of the cofficients of the unknowns, i.e. 



(12) 



A= 



ax 


6i • 


.. px 


at 


62 . 


•• Vi 


an 


K • 


•• Vn 



The determinant A is called the determinant of the system. Multiply 
the equations by ^1, — J.2, Az, — A4, etc., respectively, and add the re- 
sults. Then we have 



(13) Xi(aiAi — a^Az •••) + X2(hiA\ — h^Ai •••)+ •• 



+ ic„(pi^i-p2^2---) 
= qiAi—qzAi •••• 



From the corollary of Theorem 6 it follows that the coeflBcient of x\ is 
A and that the coefficients of the other unknowns are zero. Moreover, 
the right-hand member of (13) is the expansion of A if we replace the 
column of a's by the column of constant terms. This determinant will 
be denoted by A^^. Therefore we may write 



or 



provided A rjt 0. 
Similarly 

provided A :?t 0. 



A . iCi = Ao 



A 



Xi 



-^ 



It will be noticed that this is a direct extension of the methods employed 
in §§ 311, 313. The result may be stated in words as follows. The value of 
any unknown is equal to a fraction whose denominator is the determinant 
of the system and whose numerator is the determinant obtained from the 
former by replacing the coefficients of the unknown sought by the column 
of constant terms. 



XX, § 322] 



DETERMINANTS 



491 



321. The Case A = 0. The previous methods show that, even if 
A = 0, we can derive from the given equations the relations 

A . a:i = Aog, A • X2 = A^g, •••, A • a:« = Apq. 

Now if A = 0, th6se relations would imply that 

Aa3 = 0, A5, = 0, ..., Ap5 = 0. 

But it is easy to write down a system in which A = and one or more of 
the Aog, Aftg««-are not zero. Such a system is then clearly inconsistent 
and has no solution. For example, 2 xi + X2 = 1, 2 Xi + a;2 = 2. 

If Aag = Afeg = ••• = Apg = 0, the system may be consistent but the un- 
knowns Xi, iC2, •••, Xn are not then completely determined. For example, 
2 xi + 3^2 = 1, 4 jci + 2 X2 = 2. 

A complete discussion of this case is beyond the scope of this book.* 

322. Consistent Equations. Equations which have a common 
solution are called consistent. Consider the three equations in two un- 
knowns X and y : 

(14) aix + biy -|- ci = 0. 

(15) a2X + h-iy + C2 = 0. 

(16) azx + hzy + Cg = 0. 

Two cases arise according as to whether a pair of the three equations 
has a single or an infinite number of solutions. 

Case 1. A single solution. In order that these three equations be con- 
sistent it is necessary that 



Cl 


h 


C2 


&2 


dl 


&1 


a2 


&2 



(17) 



satisfy equation (16), i.e. that 
or its equivalent 



ai 


C\ 


«2 


C'2 


ai 


bi 


a2 


&2 



L|a2 62I J 





Cl bi 




Ol Cl 




ai 61 


- as 




-63 




+ C3 






C2 62 




(H C2 




aa &2 



= 



= 0. 



* Those interested in this problem will find a complete discussion in 
BocHEB, Higher Algebra, Chapter IV. 



492 



MATHEMATICAL ANALYSIS [XX, § 322 



Case 2. An infinite number of solutions. In this caae 

ttl _ (Z2 __ Cli 

bi 62 bs 

and hence, by the corollary of theorem 5, the above determinant must equal 
zero. Therefore, in order that three linear equations in two unknowns 
have a common solution, it is necessary that the determinant of the coeffi- 
cients of the unknowns and the known terms vanish. 

Extending this result to n linear equations in n — 1 unknowns, we 
have a necessary condition that n linear equations in n — 1 unknowns he 
consistent is that the determinant formed from the coefficients of the un- 
knowns and the knoion terms must vanish. 

It must be clearly understood that the vanishing of the above determi- 
nant is only a necessary and not a suificient condition that the equations 
be consistent. For example, the system 

2x+ y-l = 0, 
2x+ y + 5 = 0, 
4a; + 2y + 3 = 0, 
gives 

2 1-1 

A= 2 1 6 =0, 

4 2 3 

but the equations are inconsistent, for any pair are inconsistent. 



EXERCISES 

Solve the following systems of equations by means of determinants ; 



1. 



8. < 



2x — y — z = 0j 
Zx + y+z = 6, 
2x-Sy — v=—2, 
2x-hSv — d. 
— a; + ?/ + 2 = 2m, 
x-y + z=2n, 
x-\-y — z-2p. 
32/ -4x- 20 + 10 =-21, 
a; + 72/ + 2-w> = 13, 
l/-2a;-30 + 2io = 14, 
.3a; + 5^ - 50 +3wj = 11. 



4. 



x + y + w = Q, 

x + y + = 7, 

y + + w = 8, 

X + + to = 9. 

05 + 2^-0 + 3to=- 10, 

x + 3y — 20 — 4to = l, 

2x-y-30 + 5to = 3, 

3x-.y-0-2io = 18. 



XX, § 322] 



DETERMINANTS 



493 



Determine whether the following systems of equations are consistent : 

6. |a; + y-2 = 0, 7. ISx-2y + 'l=0. 8. lx-y-\-7=0, 

[4x+y-2 = 0. [3x + y-2 = 0. 



3a;-6y-2=0. 



9. Find k so that the following equations are consistent : 

2x + y-3 = 0, 
3ic-2/ = 2, 
x-^y + k = 0. 

/ 
MISCELLANEOUS EXERf^ISES 

1. Prove that the equation of the oifcle that passes through the points 
(aJi,yi), (2:2,2/2), (353,2/3) is 

(x2 + y2) X y \ 
(ici^ + yi^) a:i y-i 1 
(X2^ + y'J^) ^2 2/2 1 
(a;82 + 2/3^) a;3 2/3 1 



= 0. 



2. Prove that ax^ + 
two linear functions if 



+ 2 hxy + 2fx + 2gy + c is the product of 



a h g 
h b f 
9 f c 

3. Prove that a necessary condition that the three lines aix 4- biy + Ci 
0, a2X + 622/ + C2 = 0, asx + 632/ + ca = 0, be concurrent is that 

ai 61 Ci 

052 62 C2 = 0. 

as 63 C3 
Is this condition also suflBcient ? 

4. Prove that the locus of the equation ax-^by + = is a straight 
line. 

[Hint : Let {xi ,2/1), (icz , 2/2) be any two fixed points on the locus and 
{x, y) any other point on the locus. Then we have axi + 6yi + c = 0, 
ax2 + by2 -h c = 0, ax + by -\- c = 0. Since these equations are consistent, 
the determinant of the coefficient is zero.] 



PART V. FUNCTIONS OF TWO VARIABLES 
SOLID ANALYTIC GEOMETRY 

CHAPTER XXI 

LINEAR FUNCTIONS 

THE PLANE AND STRAIGHT LINE 

323. Introduction. Thus far the only functions which we 
have represented geometrically are those of the form y = f{x), 
i.e. functions of a single independent variable x. Such func- 
tions, in general, were seen to represent a curve in the (a;, y) 
plane. We shall now study functions of the form z = f(x, y), 
i.e. functions of two independent variables x and y. In order 
to carry out this investigation it is necessary to set up a coordi- 
nate system in three dimensions. 

324. Orthogonal Projections. The orthogonal projection of 
a point P upon a plane a (Fig. 255) is the foot P' of the per- 
pendicular drawn from P to a. The ortho- 
gonal projection of a segment PQ upon a is 
the segment PQ' joining the projections of 
P and Q upon a. 

The orthogonal projection of a point P 
upon a line I is the foot P of the perpen- 
dicular drawn from P to I. The orthogonal projection of a 
segment PQ upon the line I is the segment PQ' joining the 
projections of P and Q upon I. 

494 



XXI, § 325] 



^ 



LINEAR FUNCTIONS 



495 



325. Rectangular Coordinates in Space. Consider three 
mutually perpendicular planes intersecting in the lines X'X, 
F' F, ZiZ. These lines are themselves mutually perpendicular. 
The three planes are known as the coordinate planes and their 
three lines of intersection as the coordinate axes. The planes 
are known as the xy-plane, yz-plane, xz-plane, and the axes 
as the X-axis, y-axis, z-axis. The point which is common 
to the three planes and also to the three axes, is called the 
origin. The positive directions of these axes are usually taken 
as indicated by the arrows in Fig. 256. 




>-x 



Fig. 266 



Fig. 257 



Let P be any point in space, and let us consider the seg- 
ment OP. The numbers representing the projections of 
OP on the three axes we call the coordinates of P and 
denote them by x, y, and z. In Fig. 257, x = OA, y = OB, 

Z=:00. 

Conversely, any three real numbers x, y, z may be con- 
sidered as the coordinates of a point P. Why? If i^ is the 
foot of the perpendicular dropped from P on the xy-igleme, and 
A is the foot of the perpendicular dropped from Mon the a>axis, 
the coordinates of P are x — OA, y = AM, z = MP. 

The eight portions of space separated by the coordinate 
planes are called octants. From the preceding definitions it 



496 MATHEMATICAL ANALYSIS [XXI, § 325 

follows that the signs of the coordinates of a point P in any 
octant are as follows : 

(a) X is positive or negative according as P lies to the right 
or left of the 2/2!-plane ; 

(6) y is positive or negative according as P lies in front or 
back of the a;2;-plane ; 

(c) z is positive or negative according as P lies above or 
below the ajy-plane. 

EXERCISES 

1. What are the coordinates of the origin ? 

2. What is the z coordinate of any point in the a;y-plane ? 

3. What are the x and y coordinates of any point on the 2:-axis ? 

4. What is the locus of points for which x = ? for which y = ? 
for which ^ = ? 

5. What is the locus of points for which x = and ?/ = ? 

6. What is the locus of points for which y =0 and 2 = 0? 

7. What is the locus of points for which s = and x = ? 

8. What is the locus of points for which a; = 2 and y = 2 ? 

9. If P(x, ?/, z) is any point in space, find 

(a) its distance from the xy-plane ; (d) its distance from the x-axis ; 

(h) its distance from the ?/2;-plane ; (e) its distance from the y-axis ; 

(c) its distance from the x^-plane ; (/) its distance from the ^-axis. 

10. Describe the positions of each of the following points : (2, — 8, 3) ; 
(-2, 3, -5); (3, 3, -3); (-4, -7, -9). 

11. Plot the following points : (2, 1, 3) ; (4, ~ 1, - 2) ; (0, 0, - 3) ; 
(3,1,1); (-1,-1,-1); (1,0,1); (-1,2,-1); (1,-1,0); 
(4, - 1, - 1). 

12. Find the distance from the origin to the point P (a;, y, z). 

13. A point P moves so that its distance from the origin is always 
equal to 4. Find the equation of the locus of P. 

14. Show that the points (a;, j/, z^ and (— x, y, z) are symmetric with 
respect to the j/s-plane. 

15. A rectangular parallelepiped has three of its faces in the coordi- 
nate planes. Find the coordinates of its vertices, assuming that the di- 
mensions of the parallelepiped are a, 6, c. 



XXI, § 326] 



LINEAR FUNCTIONS 



497 




Fig. 258 



326. Directed Segments. We shall define the angle be- 
tween two directed lines I and m which do not meet, to be the 
angle between two similarly directed 
lines I' and m' which do meet (Fig. 258). 

Theorem I. 



If AB is a directed seg- 
ment on a line I, which makes an angle 6 
with the directed line V, then 

(1) Proiv AB==AB cos Q. 

Proof : Through A' (Fig. 259) draw ?i parallel to I and let 
Bi be the projection of B on l^. Then, by definition, the angle 
between I and V is the same as the angle 
between l^ and V. It follows from § 135 

*^^* A^B'==A'B, cose, 

or A'B' = AB cos By 

since A'Bi = AB. 

Theorem II. The projection on a directed line s of a broken 
line made up of the segments A1A2, AzA^, ^3^4? •••, A^^iA^, is 




equal to the projection on s of the segment AiA^. 
The proof of this theorem is left as an exercise. 



See § 136. 



Corollary. If Pi (%, yi, z^) and P^ {X2, y%y Zz) are any two 

points, then , ^ . 

'X2 - xi = Proja, P1P2, 

(2) U2 - yi = Projy P1P2, 

^2 - zi = Proj« P1P2. 



EXERCISES 

1. Find the projections upon the coordinate axes of the sides of the 
polygon ABCDEF whose vertices are A (0, 0, 0), ^ (1, - 6, 4), C (- 2, 
4, - 1), D (3, - 1, 2), E{2, 1, 4), F (1, 1, 1). 

2. The projections of the segment MP upon the coordinate axes are 
4, 3, - 1 respectively. If Jlf is (2, - 1, 3), find the coordinates of P. 

2k 



498 



MATHEMATICAL ANALYSIS [XXI, § 327 



327. Direction Cosines of a Line. Let I be any directed 
line and V a line through the origin having the same direction. 
If V makes angles a, y8, and y with the x, y, and z axes respec- 
tively, then, by definition, I makes the same angles with these 
axes. These angles are known as the direction angles of the 
line Z, while their cosines are called the direction cosines of I. 

Keversing the direction of a line changes the signs of the direc- 
tion cosines of the line. For reversing the direction of a line 
changes a, /?, y into rr — a, ir — (3, tt — y, respectively ; and by 
§122 cos(7r-^) = -cosa 

Theorem. The sum of the squares of the direction cosines of 
a line is equal to unity. 




)»X 



Fig. 260 



Proof. Let P(x, y, z) be any point on V (Fig. 260). Then, 

we have 

{x = OP cos a, 

(3) 2/ = OP cos ^, 

[2; = OF cos y. 

Therefore, 

x^-j-y'' + z^= OP^[cos2 a + cos^ jS + cos^ y]. 

Since ips -f 2/2 -f gz = Op\* it follows that 

(4) cos2 a +C0S2 p 4- cos2 7 = 1. 

* or' = x2 + y2 and OP^ = 22 -I- oS^^ = ^2 -I- xa + ya. 



± -\/P + m2 


+ 


n2 


m 






± VZ2 + 7^2 


+ 


n^' 


n 




, 



XXI, § 328] LINEAR FUNCTIONS 499 

Any three numbers /, m, n, (not all zero) are proportional to 

the direction cosines of some line ; for, P{1, m, n) is a point 

and the direction cosines of OP are 

I 
cos a = 



(5) cos (3 



cos y = 

±VZ* + m2+n2 

The direction cosines of OP are evidently proportional to Z, 
m, n, and they may be found by dividing I, m, and n, respec- 
tively, by ± V/2 + m^ 4- 7i\ 

328. The Distance between Pi(xi, i^i, zj and ^2(^2, i/2, ^2). 
Let the direction angles of the segment P1P2 be a, (3, y. Pro- 
jecting P1P2 upon the axes, we have, from the corollary of § 326, 

P1P2 cos a = .T2 — Xi, P1P2 cos /? == 2/2 — 2/ij A A cos y = 22 — 2!l. 

Squaring and adding we have, by the theorem of § 327, 

i\P2"=(x2-x,y^(y2-yiy-h(z2-z,y. 

Therefore, 

(6) P,P2 = V(X2 - X,y + (1/2 - Vlf -h (^2 - ^l)^. 

EXERCISES 

1. Find the length and the direction cosines of the segment P1P2, when 

(a) Pi is (2, 3, 4) and P2 is (- 1, 0, 5) ; 
(&) Pi is (- 1, 2, - 7), and P2 is (4, 1, 4) ; 
(c) Pi is (4, 7, 1), and P2 is (1, - 2, - 7). 

2. Prove that the triangle whose vertices are A{m^ w, p), 5(n, p, m), 
C ( jo, m, n) is equilateral. 

3. Find the direction cosines of a line which are proportional to 4, 7, 1. 

4. Find the length of a line-segment whose projections on the co- 
ordinate axes are 4, 7, 2. 



500 MATHEMATICAL ANALYSIS [XXI, § 329 

329. The Angle between Two Directed Lines. If aj, p^, y^ 

and otg) ft, 72 ^^^ the direction angles of two directed lines 
li and ^2 the angle 6 between them may be determined as 
follows. 

Draw the lines I'l and I'z through the origin, parallel to the 
given lines (Fig. 261). Then the angle between l\ and V2 is 0. 




^-x 



If P(x, y, z) is any point on l\, then, by Theorem II of § 326, 
we have ^^^.^^^ ^^ ^ p^^.^^^ OMNF, 

^'^' OP cos e = OM cos a2 + MN cos ft + iVP cos y^. 

But, 

OiJf = OP cos «!, il[fiV = OP cos ft, NP = OP cos yj. 

Therefore, 

(7) cos e = COS tti cos a2 H- cos Pi cos P2 + cos 71 cos 72- 

We shall assume that 6 is the smallest positive angle satis- 
fying equation (7). 

330. Parallel and Perpendicular Lines. If two lines are 
parallel and extend in the same direction, they are parallel to 
and agree in .direction with the same line through the origin. 

Therefore, if ai, ^1, yi and ag) ft? 72 a-^e the direction angles 
of the two lines, a^ = a^, /3i = ft, 71 = 72 ; and we may write 

(8) cos ai = cos a2, cos pi = cos p2, cos 71 = cos 72- 

Conversely, if relations (8) are satisfied, the given lines are 
parallel and extend in the same direction. Why ? 



XXI, § 331] LINEAR FUNCTIONS 501 

If the two lines are parallel but extend in opposite directions, 
we have ai = ir — 03, ft^ tt — ft? yi = 'r — 72? and therefore, 

(9) cosai=-cosa2, cos Pi=-cos pg, cos 71 =- cos 72- 

Conversely, if relations (9) are satisfied, the given lines are 
parallel and extend in opposite directions. Why ? 

If the two lines are perpendicular^ it follows from formula 
(7) that, 

(10) cos tti cos ttg + COS pi COS p2 + COS 7i cos 72 = 0. 

Conversely, if (10) is true, the lines will be perpendicular. 

If /, m, n and l\ m', n' are proportional to the direction 
cosines of two lines, the lines will be perpendicular if, and 
only if, 

(11) W + mm' + nn' = 0. 

They will be parallel if, and only if, the numbers I, m, n are 
proportional to V, m', n'. If any of the numbers I, m, n are 
zero, the corresponding numbers of the set Z', m', n' must, of 
course, also be zero. 

331. Point of Division. Let Pi(xi, 2/i> ^i), Afe Vii ^2) be 
two given points and P{x, y, z) any point on the segment P1P2 

P P 

such that ^-^ = X. If a, jS, y are the direction angles of this 

PPi 

segment, it follows from § 326 that 

PiP cos a = x — aji, PP2 cos a= X2 — x. 

Therefore, 

PiP cos a __ X — Xi __, 

PP2 cos a X2^ X 
or 

Xi +Xx2 



(12) x = 



1-hX 



502 MATHEMATICAL ANALYSIS [XXI, § 331 

Similarly, we have 

It should be noticed that A is positive if P lies within the 
segment PiPo, and negative if it lies without. By varying A, 
the coordinates of any point (=^ P2) on the line P1P2 may be 
obtained. 

For the mid-point of P1P2 we have A = 1 and, hence, the 
coordinates of the mid-point of P1P2 are 

(14) ^^ a;i-f a;2 ^ y=yi±Jh^ ^^ ^1 + 22 ^ 

i^ J 2 

EXERCISES 

1. Find the cosine of the angle between the two lines whose direction 
cosines are proportional to 2, 3, 1 and — 1, 4, 6. 

2. Find the coordinates of the points of trisection of the segment 
Pi(4, -1,3), P2(-4, 7,3). 

3. Prove that the medians of the triangle whose vertices are (1, 2, 3), 
(3, 2, 1), (2, 1, 3) meet in a point. 

4. Show that the following points are the vertices of a right triangle : 
(1,0,6), (7,3,4), (4, 5, -2). 

6. If two of the direction angles of a line are 45° and 60°, find the 
third direction angle. 

6. Prove that the values a = 30°, /3 = 30° are impossible. 

7. The direction cosines of a line are m, 2 m, 3 m. Find wi. 

8. Show that {x—\Y+{y-\- 2)2 -f. (5; _ 3)2 = 9 is the equation of a 
sphere whose center is at (1, — 2, 3) and whose radius is 3. 

9. Express by an equation the fact that the point (x, y, z) is equi- 
distant from (2, 1, 3) and (— 1, 4, 3). 

10. Show that the points (3, 7, 2), (4, 3, 1), (1, 6, 3), (2, 2, 2) are 
the vertices of a parallelogram. 

11. Prove by two methods that the points (3, 6, 4), (4, 13, 3), 
(2, — 1, 6) are collinear. 

12. Show that the points (4, 3,-4), (- 2, 3, 2), (- 2, 9, - 4) are 
the vertices of an equilateral triangle. 



XXI, § 332] LINEAR FUNCTIONS 503 

13. Find the coordinates of the point which divides the segment Pi Pa 
in the ratio X, given 

(a) Pi(2, 6, 8), P2(-l, 3, 6),X = 3; 

(6) Pi(- 2, - 5, 8), P2(8, 0, - 2), X = - 2 ; 

(c) Pi(3, - 7, - 9), P2(2, - 2, - 1), X = 1. 

14. Prove that the medians of the triangle Pi(a;i, yi, ^i), P2(X2, y2, ^2)1 
Ps(x3, ys» Z3) meet in the point 

( xi + ^2 + a;3 yi + y2 + Vs gi + g2 + ss \ 
V 3 ' 3 ' 3 / 

15. Prove that the lines joining the mid-points of the opposite edges 
of a tetrahedron pass through a common point and are bisected by that 
point. 

16. Are the following points collinear : (2, 1, 3), (—2, —5, 3), 
(1, 5, 7) ? 

17. Find the direction cosines of the line that is equally inclined to the 
three axes. 

18. Prove that the lines joining successively the middle points of the 
sides of any quadrilateral form a parallelogram. 

19. Find the projection of the segment Pi(l, 2, 3), P2(2, 1, 3) upon 
the line that passes through the points P3(— 3, 5, — 5) , P4(8, — 9, 12). 

332. Locus of an Equation. We saw that in the plane the 
locus of the equation f(x, y) =0 represents, in general, a curve. 
In an analogous way the equation f(x, y, z) = 0, in general, re- 
presents a surface. For, if we solve for z, we have z = F(x, y) 
and from this equation, we see that we can find, corresponding 
to every point (x, y) in the xy-^lame, one or more values of z 
(real or imaginary). The locus of the real points (x, y, z) is, 
in general, a surface, but may be a curve or a point. If there 
are no real values for a;, ?/, z which satisfy the equation 
f{x, y, z) — 0, we say that the equation has no locus. 

The locus of points satisfying the two conditions /(a;, y, 2!)=0 
and F{x^ y^z)—^ is, in general, a curve in space, which is the 
intersection of the two surfaces represented by these equations. 



504 MATHEMATICAL ANALYSIS [XXI, § 333 

333. The Plane. A plane is defined as a surface such that 
every point collinear with two points of the surface is itself a 
point of the surface. 

We shall prove the following propositions : 

(a) Every equation of the first degree in x, y, and z represents 
a plane. 

ih) Every plane is represented by an equation of the first de- 
gree in X, y, z. 

To prove (a), let Pj (xi, yi, z^), P^ixz, 2/2? ^2) be any two points 
on the surface whose equation is Ax -{- By + Cz -\- D = 0. 
Then we have 

(15) Ax^ + By,+ Cz^ + D^(), 

(16) Ax^ 4- By2 -h Cz^ + 2) = 0. 

Now let Pi{x^, 2/3J 2:3) be any point on the line PiP^- Then 
(if P3 ^fc Pa)) there exists a value of A.(=5fc — 1) such that 

"^-"TTT' ^^-TTT' '^"TTX" ^^^^^^ 

We wish to show that the coordinates of this point also satisfy 
the equation Ax -\- By -\- Cz + D == 0. By substitution in this 
equation we have 

(17) .^-{Ax,-\- By^-\-Cz,-\-D)-^,:^^{Ax^-^By^^-Cz2+D) = 0. 

Relation (17) is true, since it follows from (15) and (16) that 
each parenthesis vanishes separately. Therefore the surface 
defined by the equation Ax -\- By ■\- Cz -\- D = satisfies the 
definition of a plane. 

To prove the statement (6), let tt be any plane, and let OH 
be the perpendicular from which meets ir in Pj (Fig. 
262). The positive direction oi OR will be taken from to 
the plane. The direction angles of OR will be called a, )3, y 



XXI, § 333J 



LINEAR FUNCTIONS 



505 



and the length OPi will be denoted by p.* Now if P(Xj y, z) 
is any point in the plane, we have, by § 326, 




>3: 



Fig. 262 

(18) Projo50P= Tvo]onOM+ ViojonMN-^ ProJo^iVP. 

Hence the equation 

(19) X cos a 4- y cos p + z cos y = p 

is the equation of the plane. Why ? It is seen to be an equa- 
tion of the first degree in x, y, z. This form of the equation 
is called the normal form. 

It follows from the above that, if Ax -^^ By -\- Cz -\- D = i^ 
the equation of a plane, the direction cosines of a line perpen- 
dicular to the plane are proportional to A, B, C. 

It is left as an exercise to prove that to reduce Ax -\- By -\- 
C^ -f- D = to the normal form we must divide each term by 
± V^^ + B^ + C% the sign of the radical being chosen opposite 
to that oi D if D =^ 0, the same as that of C ii D = 0, the 
same as that of B if C=D = or the same as that of A if 
B=C=D = 0* 

* If the plane passes throilgh the origin we shall suppose OR is directed 
upward, and hence cos 7 >0 since 7 <7r/2. If the plane passes through the 
z-axis, then OR lies in the ccy-plane and cos7 = 0; in this case we shall sup- 
pose OR so directed that /3<7r/2 and hence cos/3>0. Finally if the plane 
coincides with the yz-plane, the positive direction on OR shall be taken as 
that on OX 





^2 




±V^o^ 


B, 


■-vc^' 


±V^2^ 




+ 02^ 



506 MATHEMATICAL ANALYSIS [XXI, § 334 

334. The Angle Between Two Planes. The angle between 
two planes is defined to be the angle between two normals {i.e. 
perpendiculars) to the planes. Let AiX -\- B^y + Cis; + Di = 
and A^x + Bay -\- C2Z -{- D2 = ^ be the equations of the two 
planes. The direction cosines of their normals are then (§ 333), 

cos «! = ^ , cos 02 = 

cos Bi = ^ cos ^2 = 

cos yi = ^ ; cos y2 = 

If ^ is the angle between these normals, then, from § 329, 

(20) cos ^ = ± A,A 2±BA + C,C2 

V^l^ + B{^ + Ci2 V^z^ + ^2^ + C22 

If the planes are perpendicular, cos ^ = 0, and we have 

(21) A^A^ + B1B2 4- Ci(72 = 0. 

If the planes are parallel their normals are parallel. Hence, 
by § 330, their equations in normal form are 

X cos a-\-y cosyS-j-g cos y—p, x cos «'+ ?/ cosy8'+2;cos y'=p', 

where either cos a = cos a', cos ft = cos ft', cos y = cos y', or 
cos a = — cos a', cos y8 = — cos ft', cos y = — cos y'. Therefore, 
if the two equations be written in the form 

A^x + Biy + Ciz + Di = 0, AzX + ^22/ + C^z + A = 0, . 

the planes will be parallel if and only if 

(22) A2 = kA,, B2 = kB,, Cz^kCi. (k^O) 

The equation of any plane parallel to Ax -\- By + Cz -\- D = 
can therefore be written in the form Ax -^ By + Cz + D' = 0. 



XXI, § 334] LINEAR FUNCTIONS 507 

EXERCISES 

1. Sketch the planes whose equations are (a) a; =2, (6) y = 4, 
{c) z=-6,(id) 2x + y = l, (e)y-z = 0. 

2. How many arbitrary constants are there in the equation of the 
plane Ax+By+Cz + D = 0? 

3. What is the general equation of a plane that passes through the 
origin ? 

4. What is the equation of the x^z-plane ? y«-plane ? a;2!-plane ? 

5. What are the intercepts on the axes of the planes whose equations 
are 

(o) 2x-3y-\-z = 12', (6) x-y+z=8; (c) x + y = 0; (d) 6x-7=0? 

6. Give three numbers proportional to the direction cosines of the 
normal to the plane x-\-2y — z = 9. What are the direction cosines ? 

7. What is the normal equation of the plane x — y + z = 9? 

8. What is the equation of the system of planes parallel to 

2x — y + 2! = 1? 

9. What is the equation of the plane that passes through the origin 
and is parallel to 2x — Sy + 7z = b? 

10. Show that the planes 2x + 4ty — z = 2 and 4a;— y-f40 = 7 are 
perpendicular. 

11. What is the equation of the plane parallel to2x + 2y-\-z=:9 and 
6 units farther from the origin ? 2 units nearer ? 

12. What is the distance between the parallel planes 2x-i-2y + z = 9 
and 2 X + 2 ?/ + « = 15 ? 

13. Find the equation of the plane passing through the points 

(a) (1,2,1), (-1,1,0), (0,0,1); 
(&) (2,1,3), (1,1,2), (-1,1,4); 

(c) (2,2,2), (1,1, -2), (1,-1,0); 

(d) (1,1,-1), (1,-1,2), (-2,-2,2). 

[ Hint : Use the equation Ax-\-By+Cz + I)=0 and divide by any coeffi- 
cient that is not zero.] 

14. If D 9^ show that the equation Ax -\- By + Cz ■}- D = can be 
written in the form x/a-{-y/b + z/c = l where a, b, c are the inter- 
cepts made by the plane on the x, y, z axes respectively. 

15. Show that the four points (0, 0, 3), (4, —3, —9), (2, 1, 2), 
(4, 3, 3) are coplanar, i.e. they lie in the same plane. 



508 MATHEMATICAL ANALYSIS [XXI, § 334 

16. Find the equation of the plane that passes through the point P 
and is parallel to the plane a, when 

(a) Pis (2, 1, 8) and ais2£c + 3y- 50 = 5 ; 

(6) P is (1, 0, 0) and a is 2« + 2/ + = 1 ; 

(c) Pis (-2, - 1,6) and a is 3a;- 5 2/- 2« = 3. 

17. Find the equation of the plane passing through the point P and 
perpendicular to the planes a and /3 when 

(a) P is (1, 1, 1), a is 2 x — 2/ — = 4, and ^isx-^y + z = l; 

(b) Pis (-1, 2, 1), a is a; + 2/ -3« = 3, and /3 is .Sx- 5y + 22 = 1 ; 

(c) Pis (0, 3, 4), ais2a: + 4y + = 7, and /3is2x-0 + 32! = 2. 

18. Find the equation of the plane passing through the points Pi, Pa 
and perpendicular to the plane a, when 

(a) Pi is (1, 1, 1), P2is (-1,2, 1), and a is 2x- 3y - ;s = 2; 

(b) Pi is (0, 0, 1), P2 is (2, 1, 3), and aisx + y — 6;s = 0; 

(c) Pi is (2, 1, - 3), P2 is (0, 4, 2), and «is4x — y — ^ = 2. 

19. Prove that the distance from the plane Ax + By + Cfe + D = to 
the point (XI, 2/1, 01) is + Ax, + By,+ Cz, + D^ 

VA^ + P2 + C=2 

10. Find the distance from the plane a to the point P when 
(a) P is (2, 1, 4) and ais2x — iy -}■ z = 2 ; 
(6) Pis (2,3, - 1) and a is 2x + 2/ + 260-2 = 0; 
(c) P is (0, 0, 3) and ais3x-22/-50 = l. 

21. Prove that the equation of the plane which passes through the 
point (xi, 2/1, «i) and is parallel to the plane Ax + By -\- Cz -{■ D = is 
A(x - xi) + B{3j - 2/1) + C{z - zi) = 0. 

22. Prove that the equation of a plane which passes through the 
point (xi, 2/1, zi) and is perpendicular to the plane Ax+By+Cz+D = 
is (P01— C2/i)x+ (Cxi— ^01)2/+ (^2/1 — Bxi)z = 0. 

23. Find the cosines of the angles between the following pairs of planes 

(a) 2x-32/ + = l, 2x + z=z0; 
(6) x-y-0 = 2, 2^-40 = 8; 
(c)x + = 3, 4x + 2/+30 = 5. 

24. Find the equation of the plane that passes through Pi, P2 and 
makes an angle with the plane a, where 

(a) Pi is (0, -1, 0), P2 is (0, 0, -1), a is 2/+ 0-7 = 0, and d is 120°; 
(6) Pi is (1, 0, 1), P2 is (0, 1, 2), a is X + 2 2/ + 20 = 2, and 6 is 60°. 



XXI, § 335] LINEAR FUNCTIONS 509 

25. Find the equation of the locus of a point which moves so that its 
distance from the x?/-plane is twice its distance from the a^-axis. 

26. Find the equation of the locus of a point whose distance from the 
plane x + 2y — 5 = 0is twice its distance from the z-a,xis. 

27. A point moves so that its distance from the origin is equal to its 
distance from the ^rx-plane. Find the equation of its locus. 

335. Simultaneous Linear Equations. In § 70 we saw that 
three simultaneous linear equations in three unknowns have 
in general a single solution. We shall now show that three 
such simultaneous equations have either, (a) a single solu- 
tion, or (6) an infinite number of solutions, or (c) no solution. 

We shall prove this statement geometrically. Each equa- 
tion represents a plane ; the three planes may assume the fol- 
lowing relative positions. 

Case I. No two of the planes are parallel or coincident. 

(a) The three planes may intersect in a single point ; then 
there is a single solution of the three simultaneous equations. 

(&) The three planes may intersect in a line ; then there is 
an infinite number of solutions. 

(c) The three planes may intersect so that the three lines of 
intersection are parallel ; then there is no solution. 

Case II. Two of the planes are parallel but not coincident. 

In this case the three planes can have no point in common 
and the equations have no solution. 

Case III. Two of the planes are coincident. 

(a) The third plane may be parallel to the coincident planes, 
in which case there is no solution. 

(6) The third plane may intersect the coincident planes, in 
which case there is an infinite number of solutions. 

(c) The third plane may coincide with the coincident planes, 
in which case there is an infinite number of solutions. 



510 MATHEMATICAL ANALYSIS [XXI, § 336 

336. Pencil of Planes. All the planes that pass through a 
given line are said to form a pencil of planes. If 

A,x + B,y + Ciz H- A = 0, 



(23) 

^ [A2X + B^y + C^z -f A = 0, 

are the equations of any two planes passing through the given 
line, then the equation of any other plane of the pencil can be 
written in the form 

(24) A^x + B^y j^C,z+D,+\ {A^x -h B^ + C<,z + A) = 0, 

where A is a constant whose value determines the particular 
plane of the pencil. (See § 68.) 

337. Bundle of Planes. All the planes that pass through a 
common point are said to form a bundle of planes, and this 
common point is called the center of the bundle. If 

' A^x + Biy+C^z + D^ = 0, 

(25) A2X + A2/ + C2Z + A = 0, 
A^x + B,y^C^z+ A-0, 

are the equations of any three planes passing through the 
center and not belonging to the same pencil, then the equation 
of any other plane of the bundle is 

(26) {A^x + Biy + C^z + D^) + X^A^x + B^y + C^z + A) 

+ X2{Ax 4- B,y + C> + A) = 0, 
where Ai, A2 are constants whose values determine the position 
of the particular plane of the bundle. Why ? 

EXERCISES 

1. Find the equation of the plane that passes through the Intersection 
of the planes a and /3 and the point P, when 

(a) ais2x + 3y--2; = l, /3isa; + 2/-2« = 2, and P is (1, 0, 2) j 
(6) aisx + y + 2^ = 0, /3is4a;-2?/-2; = l, and Pis (2, 1,1); 
(c) a is 3 a: - 2 y - 2r = 2, /3 is X - 2/ + « = 3, and P is (1, 0, 1). 



XXI, § 338] LINEAR FUNCTIONS 511 

2. Show that the planes whose equations are 3a;--5y + 2 = 0, 6x + 
y = 2 2 + 13, lly — 2z = 17, belong to the same pencil. 

3. What is the equation of the plane of the pencil whose axis is 
2x — y + 52! + 2 = 0, 4x — Sy-\-z = l, which is perpendicular to the 
plane x = 0? y = 0? z = 0? 

4. Find the equation of the plane that passes through the intersection 
of the planes 2x + y — z -j- 1, Sx— y — z = 2 and is perpendicular to the 
plane x -^ y — z = 1. 

5. Find the equation of the plane that passes through the point of in- 
tersection of the planes a, /3, y and the points Pi, P2, when 

(a) ais2x + y = 1, fi is x — z = 1, yis2x — y + 2z = 3, 
Pi is (1, 0, 1), and P2 is (2, 1, 1) ; 

(b) aisSx- y -z = S, ^isx — y-\-2z = l, VisSx — 2y + 2! = 3, 
Pi is (2, 1, 3), and P2 is (0, 8, 0). 

338. Equations of a Straight Line, (a) The two simul- 
taneous equations 



(27) 

^ ^ ^ A2X + B^y-h C2Z -}- A = 0, 

represent a line, the intersection of the two planes, provided 
the two planes are not parallel. 

(6) A given point and a given direction determine a line. 
Let the given point be Pi{xi, y-^, z^ and a, /8, y the given direc- 
tion angles. If P (x, y, z) is any other point on the line at a 
distance d from Pj, then by § 326, dcos a=x—Xijd cos /8=2/— 2/ij 
d cos y = z — Zi. Hence we may write 

(28) x-x^ ^ y-yi ^ z- z^ ^ 

cos a cos p cos y * 

which are the equations of the required straight line. These 
equations are known as the symmetric equations of a straight 
line. In these equations cos a, cos /8, cos y can evidently be 
replaced by ar.y three numbers proportional to them. 



512 MATHEMATICAL ANALYSIS [XXI, § 338 

(c) Two distinct points Pi(xi,yi, Zi), P^ixz, 2/2? %) determine a 
line. Any line through, the point Pj is of the form 

cos a cos ^ cos y 

Now the direction cosines of PiA are proportional to a^ — ajj, 
?/2 — J/i> 2^2 — 2:1. (§ 328.) Therefore the equations of the line 
through the points Pi, P.^ are 

(29) x-Xi ^ y -Vi ^ z-Zi 

X2 -Xi Vi- Vi ^2 - 2i 

We should note that in every case two equations are necessary 
to represent a line. 

Example 1. Reduce to the symmetric form the equations of the straight 

line, 2a: + y--0 = 3, x — y+2z-=:1. Eliminating y between the two 

equations we have 3 a; + = 10. Similarly, eliminating z we have 

6 aj -f y = 13. Solving these two equations for x and equating the values 

found, we have 

a;_ y — 13 _ g-10 

1~ -6 ~ -3 * 

The line is seen to pass through the point (0, 13, 10) and to have direction 
cosines proportional to 1, -- 5, — 3. 

Example 2. Find the equations of the line that passes through the 
point (4, —1, 3) and is perpendicular to the plane 2aj — 3y + 40=7. 
The required line is parallel to any line perpendicular to the plane and hence 
its direction cosines are proportional to 2, —3, 4 (§333). Therefore, 
the equation of the required line is 

a;~4 _ y + l _ g-3 
2 -3 ~ 4 * 



EXERCISES 

1. Write the equations of the line that passes through the point P 
and whose direction cosines are proportional to a, 6, c, where 

(a) Pis (1, 2, 1) and a = 2, 6 = - 7, c = 2; 
(6) P is (3, 0,-1) and a -2, 6 = 8, c = 9; 
(c) P is (3, - 2, -6) and a = 2, 6 = - 9, c = 3. 



XXI, § 338] LINEAR FUNCTIONS 513 

2. Find the equations of the lines passing through the following pairs 
of points : 

(a) (2, 1, 4), (12, 2, 8) ; (c) (4, 3, 8), (8, - 2, 1) ; 

(6) (3, - 6, - 3), (- 3, 5, 7) ; (d) (5, 2, 1), (4, 7, -9). 

3. Write in symmetric form the equations of the lines 

(a) 2x-7j + Sz = S, Sx + by -^z=:9; 

(b) Sx-y-z = 8, 4:X + Qy-5z = S; 

(c) 5x-\-Sy + z=S, 2x — y + z = 'l. 

4. Find the equations of the line that passes through the point P and 
is perpendicular to the plane a, when 

(a) Pis (2, 1, 7) and ais3ic-?/ + 40 = 9; 

(6) P is (4, 2, - 2), and a is 2 a; - 6 2/ + 3 = 3 ; 

(c) Pis (— 1, 6, 3), and a is 3a; + 4?/ — ;s = 5. 

6. Find the equations of the line that passes through the point 
(2,— 1, 4) and is parallel to the line 

x-_3_ y-7 _ g-7 
4 ~ 2 -3 ' 

6. Find in symmetric form the equations of the line that passes 
through the point (2, — 1, 4) and is parallel to the line 2x + y — z = 6, 
X — y + Zz = 4:. 

7. Find the equation of the plane that passes through the point P and 
is perpendicular to the line Z, when 

(a) Pis(2,5, 1), andns^ = ^li = ^; 
o o o 

(6) Pis (- 1, 4,7), and I is =^ = 1^ = ^-±f. 

8. Find the equation of the plane that passes through P(l, 5, 2) and 
is perpendicular to the line 3x-y+2 = 8,x — y + 22! = 6. 

9 If tf is the angle between the two lines ^ ~ ^^ = V^rJll - lull^ 

«2 62 C2 

10. Prove that the lines - = -^ = -^ , ~ = ^ = - are perpendicular 
to each other. 6-2-4463 



2h 



CHAPTER XXII 

QUADRATIC FUNCTIONS. QUADRIC SURFACES 

339. The Sphere. If a point P{x, y, z) moves so as to be 
always at a constant distance r (r > 0) from a fixed point {h, k, I), 
the locus of P is called a sphere. The equation of this locus is 

(1) (x-hy+(y-ky-{.(z-iy = r^. 

If this equation is expanded, it has the form 

(2) x^-\-y^-\-z^-\-Ax-{-By + Cz-^D = 0, 

where A, B, C, D are constants depending upon the coordinates 
of the center and the length of the radius. 

Conversely, an equation of the form (2), in general, repre- 
sents a sphere, for it can be written in the form 

<" ('+f)'-('-f)'H'+f)'=t'-f-?-^- 

which is a sphere if 

4 4 4 

The center of the sphere is at the point (— A/2, — B/2, — 0/2), 
and the radius is 



V^V4 + S2/4 + 074 - D. 

If the right-hand member of (3) is zero, the locus is the 
single point (— Ji/2, —B/2, — (7/2). If the right-hand member 
of (3) is less than zero, the equation has no locus. See § 206. 

614 



XXII, § 340] QUADRIC SURFACES 515 

EXERCISES 

1. Find the equation of the sphere whose center is at P and whose 
radius is r, when 

(a) Pis (2, 1, 9),andr = 6; 
(6) Pis(l, -8,0), andr = 2; 
(c) P is (4, -9,-2), and r = 7. 

2. Find the equations of the eight spheres tangent to the three co- 
ordinate planes and having a radius of 4, 

3. Find the equation of the sphere which has the line joining P(2, 6, 8) 
and ^(4, 6, 6) as a diameter. 

4. Discuss the locus of each of the following equations, 
(a) x2 + ?/2 + ^2 _ 2 X - 2 y - 2 2 = 6 . 

(ft) a;2 + y2 _}. 2.2 _|. 4 a; __|_ 4 y _ 6 2 + 25 = 0. 
(c) a;2 + 2/2 + ^2 _ 2 X - 6y + 8 = 5. 
{d) x2 + 2/2 -f ;22 _ 2 2 - 4 y + 5 = 0. 

5. Find the locus of points the ratio of whose distances from (0, 1, 0) 
and (1, 2, 3) is 5. 

6. Show that the equation of the tangent plane to the sphere 

x'i + y^^Z^ =^2 

at the point (xi, j/i, zi) is 

xxi -f yyi + zzi = r2. 
[Hint : The tangent plane is perpendicular to the radius.] 

7. Find the equation of the sphere passing through the following 
four points. 

(a) (1,2,3), (3,1,0), (2,1,0), (3,4, 1). 
(6) (2, 1, 0), (- 1, - 1, 0), (3, 0, 2), (0, 0,0). 

[Hint : Use the equation x^ -h y^ -^ z'^ + Ax -]- By -\- Cz + D = and 
determine the values of A, J5, O, Z>.] 

340. Cylinders. The surface generated by a straight line 
which moves parallel to a given line and always intersects a 
given fixed curve, is called a cylindrical surface or a cylinder. 
The generating line in any of its positions is called an element 
of the cylinder. 

Any algebraic equation in two cartesian coordinates repre- 
sents in space a cylinder whose elements are parallel to the 
axis of the third variable. 



516 MATHEMATICAL ANALYSIS [XXII, § 340 

For example, the equation 

a;2 + y2 =4 

represents in the ccy-plane a circle (Fig. 263). But, the equation is satis- 
fied by the coordinates of any point P which lies on a line parallel to the 




«r 



2;-axi8 and passes through a point Q on the circle. Moreover, if QP moves 
parallel to the 2-axis and continues to cut the circle the coordinates of P 
still satisfy the equation x^ + y2 _ 4, The cylinder traced by the line 
QP is the locus of the equation x^ + 2^2 _ 4, 

It is clear that if a cylinder has its axis parallel to a coordinate axis, a 
section made by a plane perpendicular to that axis is a curve parallel 
and equal to the directing curve on the coordinate plane. Thus the 
section cut by the plane z = S from the hyperbolic cylinder whose equa- 
tion is 

x^-y^ = 4, 

is a hyperbola equal and parallel to the hyperbola in the xy-plane whose 
equation is x^ — y^ = 4. 

341. The Projecting Cylinders of a Curve. A cylinder 
whose elements are parallel to one of the coordinate axes and 
always intersect a fixed curve in space, is called a projecting 
cylinder of the curve. The equations of the projecting cylin- 
ders may be found by eliminating in turn each of the variables 
X, y, z, from the equations of the curve. Why ? The curve 
may often be constructed conveniently by means of two dis- 
tinct projecting cylinders. 



XXII, §342] QUADRIC SURFACES 517 

EXERCISES 

1. Describe the locus of each of the following equations, 
(a) aj = 2. (A) yz - 6. 

(6) 2x2 + y2 = 8. (4) ^_^ = o. 

^^^ ^^ ^' (m) y2==3c2. 

(fir) 2/2 - ig2 = 1. 

2. Prove that x^ + 2xy +y^ = 1 — z^ is the equation of a cylinder, 
the direction cosines of any element being proportional to (1, 1, 0). 

3. Find the equations of the projecting cylinders of each of the following 
curves. Construct the curves as the intersection of two of these cylinders. 

(a) x2 + y2 + 2,2 = 4^ a;2 + y2 _ 5,2 = 0. 
(6) a; = 1, x2 + y2 4. ;5,2 = 4. 
{c) x^ - y^ = ^ Zj x^ + y^ = z. 
{d) y'2 = x + z,z = x + yK 
(e) 02 = xy, x2 = yz. 

342. Symmetry, Intercepts, Traces, Sections. If a given 
equation is unaffected by replacing x hj — x throughout, the 
locus is symmetric with respect to the yz-^lane. 

If a given equation is unaffected by replacing y by — y, the 
locus is symmetric with respect to the a;2;-plane. 

If a given equation is unaffected by replacing zhj — z, the 
locus is symmetric with respect to the xy-i^\sine. 

What would be a test for symmetry with respect to the a;-axis ? the 
t/-axis ? the «-axis ? the origin ? 

The segments measured from the origin to where a surface 
cuts the axes are called the intercepts of the surface on the 
axes. To find the intercepts place two of the variables equal 
to zero and solve the resulting equation for the third variable. 
Why? 



518 



MATHEMATICAL ANALYSIS [XXII, § 342 



The sections of a surface made by the coordinate planes are 
called the traces of the surface (Fig. 264). To find the equa- 
tions of the traces put each variable in 
turn in the given equation equal :to zero. 
Why? 

The equations f(x, y, z)—0 and x = k, 

a constant, are together the equations of 

the curve of intersection of the surface 

and a plane parallel to the i/^^-plane. 

Similarly sections parallel to the xy- and 2/z-planes may be 

found. If A: = 0, the sections are the traces. 




*-x 



343. The Ellipsoid. The surface represented by the equation 



w 



1' 4.1^' 4.^=1 
a-i b^ c2 



is called a,ii- ellipsoid. It is symmetric 
with respect to the three coordinate 
planes, the three axes, and the origin. 
The intercepts on the x-, y-, s-axes are 
respectively ± a, ±b, ± c (Fig. 266).* 
The traces on the three coordinate planes 
are, respectively. 



x^_^y 



&2 



1, . = 0; ^' + ?-!=l, 2/ 



0; 







Fig. 265 



1, X = 0. 



The sections of the ellipsoid by the plane a; = A; is an ellipse 
whose equations are 



('-'f<'-S) 



= 1, xz=k. 



bVl 



* The figure exhibits only that part of the surface lying in one octant,— 
that in which x, y, z are all positive. 



XXII, § 343] 



QUADRIC SURFACES 



519 



The semi-axes of this ellipse are 6 VI — W-ja^, cVT-^F/o^. 
As I A; I increases from to a, the axes of this elliptical section 
decrease. When | A; | = a the ellipse reduces to a point, and 
when I A; I > a the sections are imaginary. The surface lies 
therefore entirely between the planes x — a^ x = — a. Sim- 
ilarly it may be shown that the surface is also bounded by the 
planes y = b, y — — b ; z = c, z = — r. 



LI 




' 




n 


--^ 



Fig. 26() 

A general idea of the appearance of an ellipsoid is given 
by Fig. 266, which represents a plaster model of this sur- 
face. 

Special Cases. In general the semi-axes a, b, c are unequal, 
but it may happen that two or three of them are equal. If 
the three are equal, i.e. a = b = c, the surface is a sphere. If 
two are equal, for example, if & = c, the ellipsoid is called an 
ellipsoid of revolution, for it can be generated by revolving the 
ellipse x'^/a^ -\- y^/b^ = 1, j; = about the avaxis. 




520 MATHEMATICAL ANALYSIS [XXII, § 344 

344. Surfaces of Revolution. The surface generated by 
revolving a plane curve about a line in its plane is called a 
surface of revolution. The equation of the surface is readily 
found when the axis of revolution, i.e. the line about which 
the curve is revolved, is one of the coordinate axes. 

Let y=f{x) be the equation of the 
plane curve in the a;?/-plane and the 
ic-axis the axis of revolution. As the 
curve 2/= /(a?) revolves about the a;-axis, 
any point P on this curve describes a 
circle, whose center is on the x-axis and 
whose radius is equal to f{x) (Fig. 267). 
Therefore for any position of P (x, y, z) we have, 

which is the equation of the required surface of revolution. 

If the ellipse rK'^/a^ -f y^/h^ = 1, ^ = is revolved about the x-axis, the 
equation of the surface of revolution is 

^^ + ^^ = ita^-^^3, or 1 + 1^,4-^=1. 

EXERCISES 

1. Sketch and discuss each of the following ellipsoida. 
(a) 9 a:2 + 4 2/2 + 16 z-^ = 144. 

(6) 25 a;2 + y2 _i_ ;j;2 =: loo. 

(C) x2 + 8y2 + 2 2;2=:16. 

2. Show that the ellipsoid in Ex. 1 (6) is an ellipsoid of revolution. 

3. Find the equations of the ellipsoids formed by revolving the follow- 
ing ellipses about the axes mentioned. 

(a) 9 x2 + 4 y2 = 36, « = o, x-axis. 

(6) 9 x2 + 4 y2 = 36, « = 0, 2/-axis. 

(c) 9 x2 + «2 = 9, ?/ = 0, ^-axis. 

(d) 25 2/2 + 4 22 = 100, X = 0, ?/-axi8. 



XXII, § 345] 



QUADRIC SURFACES 



521 



4. When an ellipse is revolved about its major axis the ellipsoid gen- 
erated is called a prolate spheroid; when it is revolved about its minor 
axis, an oblate spheroid. Which of the ellipsoids in Ex. 3 are oblate and 
which are prolate ? 

5. Describe the locus of each of the following equations. 



(a) ^ + ^ + 5_ = o 
^ ^ a^ b^ c'' 



^ ^ a^ b^ d^ 



345. The Hyperboloid of One Sheet. The surf ace. repre- 
sented by the equation 



(6) 



Qi "^ 62 ^2 



is called a hyperboloid of one sheet 

respect to each of the coordinate 
planes, each of the coordinate 
axes, and the origin. The inter- 
cepts on the X- and ?/-axes are 
± a and ± b respectively, while 
the surface does not meet the 
z-axis (Fig. 268). The traces on 
the coordinate planes are, respec- 
tively, 



It is symmetric with 







Fig. 268 



--- = 1, 2/=0. 

a2 c2 ' ^ 



Of these, the trace on the xy-^lsme is an ellipse, while the other 
two are hyperbolas. 

The section of the surface made by the plane z = k, is an 
ellipse whose equations are 



•t'?]''t'n 



= 1, z = A;. 



522 



MATHEMATICAL ANALYSIS [XXII, § 345 



This ellipse is real for all real values of k. The semi-axes are 
the smallest when A; = and increase without limit as | A: | 
increases. 

The plane y = X, | A. | ^ 6, intersects the surface in the 
hyperbola 

= 1, 2/ = X. 



•■['-^] I'-s] 



If I A I < 6 the transverse axis is parallel to the ic-axis, while 
if I A I > 6 it is parallel to the 2;-axis. 




Fig. 209 



A good idea of the appearance of this surface is given by 
Fig. 269, which represents a plaster model of a portion of the 
surface. 

If A, = 6, the section consists of the two straight lines 



a c 



a c 



XXII, § 345] QUADRIC SURFACES 523 

If A. = — 6, the section is the two lines 

a c a c 

These four straight lines lie entirely upon the surface. 

Similar considerations apply to the sections made by planes 
parallel to the yz-^lsme. 

. The form of one eighth of the surface is given in Fig. 268. 
The broken lines in that figure indicate three sections by the 
three planes 

y = \, for I A, I < 6, =b, and > b. 

Some of the straight lines on the surface are shown on the 
model represented by Fig. 269. 

If a = 6 the hyperboloid becomes a surface of revolution 
obtained by revolving the hyperbola x^/a^ — z^/c^ = 1, y =zO 
about its conjugate axis. 

EXERCISES 

1. Sketch and discuss each of the following surfaces. 

(a) x2+4 2/2_g2=16. (ft) 9x^-\-y^-z^=Se. (c) ix^+16y^-z-^=Qi. 

2. Are any of the surfaces in Ex. 1 surfaces of revolution ? 

3. Show that 

(x-2y (y-l)2 (z-Sr ^^ 
9 4 1 

is the equation of a hyperboloid of one sheet whose center is at the point 
(2, 1, 3). 

4. Show that ^-t: + ?l = i and -^ + l^+£!=3l are equations of 

a'^ b'^ c^ a'^ 52 ^2 

hyperboloids of one sheet. 

5. Find the equation of the hyperboloid of revolution formed by 
revolving each of the following hyperbolas about the axis specified. 

(a) 9 a;2 - 4 1/2 = 36, z = 0, transverse axis. 
(6) 9 x2 - 4 2/2 = 36, ;j; _ 0, conjugate axis. 

(c) 4 1/2 _ 2;2 _. 16, a; = 0, transverse axis. 

(d) 4 1/2. _ z2 = 16, X = 0, conjugate axis. 



524 



MATHEMATICAL ANALYSIS [XXII, § 346 



346. Hyperboloid of Two Sheets. The surface represented 
by the equation 

(6) 



fl2 e,2 



22 



C2 



==1 



is called a hyperboloid of two sheets. It is symmetric with 



^x 




Fig. 270 



respect to each of the coordinate planes, the coordinate axes 
and the origin. The intercepts on the a?-axis are ± a, while 




Fig. 271 



the surface does not meet the y- or z-axis (Fig. 270). The 
traces on the coordinate planes are, respectively. 



a;2 y' 



There is no trace on the 2/2?-plane. 






XXII, § 346] QUADRIC SURFACES 525 

The plane a; = A; intersects the surface in the curve whose 
equations, iik^± a, are 



11-'] {5-] 



1, x = k. 



If I A: I > a this curve is an ellipse ; if \k\ = a it is a point. 
If 1 A: I < a the equations have no locus. All sections parallel 
to the xy- and a;2;-planes are hyperbolas. 

A good idea of the appearance of this surface is given by Fig. 
271, which represents a model of a portion of the surface. 

If 6 = c the hyperboloid becomes a surface of revolution 
formed by revolving the hyperbola 

X- 






about its transverse axis. 



EXERCISES 

1. Construct and discuss each of the following surfaces, 
(a) 4 a;2 ~ 9 ^2 _ 36 ^2 = 144. 

(6) x2 - y2 _ z2 - 1. 
(c) 9x2-4«/2-02 = 36. 

2. Are any»of the surfaces in Ex. 1 surfaces of revolution ? 

3. Show that _ ±- + ^ - 5- = 1 and _ ±- _ ^^ + ^ = 1 are equations 

a2 &2 c2 02 62 ^c2 ^ 

of hyperboloids of two sheets. 

4. Find the equation of the hyperboloid of revolution formed by 
revolving each of the following hyperbolas about the axis specified. 

(a) ^ - ^ = 1, = 0, conjugate axis. 
4 9 

(6) 4:y^ - z^=4, x = 0, transverse axis. 

(c) 2 x2 — 4 02 = 1^ y = 0, conjugate axis. 



526 



MATHEMATICAL ANALYSIS [XXII, § 347 



347. The Elliptic Paraboloid. The surface represented by 
the equation 



is called an elliptic paraboloid. It is symmetric with respect 
to the xz- and 2/2j-planes, and the 2-axis. The 
intercepts on all three axes are zero. The trace 
on the a;y-plane is a point, namely the origin ; 
the traces on the xz- and yz-iplsmes are, respec- 
tively, the parabolas x^ = a^z, y = 0; y^ = b^z, 
x = (Fig. 272). 

Sections made by the planes z = k (k > 0) 

are ellipses. Why ? Those made by the planes x = k and 

y = kj respectively, are parabolas. Why ? 




Fig. 272 




FiQ. 273 



Figure 273 represents a model of a portion of the surface. 
If a = 6, the surface is a figure of revolution formed by re- 
volving the parabola x^ = a% y = 0, about the 2;-axis. 



XXII, § 348] 



QUADRIG SURFACES 



527 



348. The Hyperbolic Paraboloid. The surface represented 
by the equation 

x2 y2 



(8) 



fe2 



is called a hyperbolic paraboloid. (See Fig. 274.) It is sym- 
metric with respect to the xz- and yz-iplsmes. All three inter- 
cepts are zero. The trace on the xy-iplsme is the pair of lines 



^ . ?/ 



0, ^=0; 



the traces on the xz- and t/^-planes are, respectively, the 
parabolas 

x^ = a% y = 0; y^ = - b% a; = 0. 





Fig. 274 



Fig. 275 



Sections parallel to the a;?/-plane are hyperbolas, while those 
parallel to the xz- and 2/2!-planes are parabolas. The form of 
the surface is shown in Fig. ?75. 



528 MATHEMATICAL ANALYSIS [XXII, § 348 

EXERCISES 

1. Sketch and discuss each of the following surfaces, 
(a) a;2 + 4y2 = 36 z. (c) y'^ - z^ = x. 

(6) 2X2 + ;32 = 16y. (d) 2X2 - 02 = _ y. 

2. Sketch the surface «2-2a; + t^-4y = «~5. 

349. The Cone. The surface represented by the equation 

is called a cone. It is symmetric with respect to the three 
coordinate planes, the three axes, and the origin. All three 
intercepts are zero. The trace on the xy- 
plane is a point, namely the origin. The 
traces on the x%- and 2/2!-planes are respec- 
tively the pairs of lines ca; ± az = 0, 2/ = ; 
c?/ ± 6^ = 0, a; = (Fig. 276). Sections paral- 
lel to the a;?/-plane are ellipses, while those 
parallel to the xz- and 2/2!-planea are hyper- 
bolas. If any point P (aji, ?/i, z^) on the sur- 
face is connected with the origin, then the line OP lies entirely 
on the surface. For, (Xaji, Xa:2, X%) are the coordinates of any 
point on this line (see § 331), and they arc seen to satisfy the 
given equation (9), for all values of X. 
If a = 6 the cone is a cone of revolution. 

EXERCISES 

1. Construct and discuss each of the following surfaces. 

(a) x2 + t/2-;22 = o. (6)9x2 + 4 2/2-3602-0. (c) x2-y2 4.4;s2 = o. 

2. A point P moves so as to be equidistant from a plane and a line 
perpendicular to the plane. Find the equation of the locus of P. 

3. A point P moves so that the sum of its distances from the three 
coordinate planes is equal to its distance from the origin. Find the 
equation of the locus of P. 




XXII, § 350] QUADRIC SURFACES 529 

350. Summary. The surfaces discussed are here enumer- 
ated for reference. 
Ellipsoid : 

1 + !i + -2=l- (Figs. 265, 266, § 343) 

w- 0^ c^ 

HyPERBOLOID OF ONE SHEET: 

-! + S - ^ = ^- (^^^«- 2^^' 269, § 345) 

0} ©2 Qi 

Hyperboloid of two sheets : 
a;2 ?/2 



1. (Figs. 270, 271, § 346) 



Elliptic Paraboloid : 



a2 62 
Hyperbolic Paraboloid : 

/2 



t -I- 1 = z. (Figs. 272, 273, § 347) 



^-|^ = ^. (Figs. 274, 275, § 348) 

a 



Quadric Cone: 



S + S-| = 0- (Kg. 276, § 349) 



QuADRic Cylinders : 
a;2 



±|-2 = 1, y^ = ^px, (§340) 



It is beyond the scope of this book to prove that the general 
equation of the second degree in three variables x, y, z, can, in 
general, be reduced to one of the above types. Those inter- 
ested in this problem will find it fully discussed in any stand- 
ard textbook on solid analytic geometry.* 

*See, for example, Snyder and Sisam, Analytic Oeometry of Space, 
Chapter 7. 

2m 



530 



MATHEMATICAL ANALYSIS [XXII, § 351 




y 



351. Other Systems of Coordinates. 

Numerous systems of coordinates for deter- 
*'^ mining the position of a point P in space have 
been devised. The most common of these sys- 
• ^^ tems are the rectangular, polar, spherical, and 

cylindrical. A brief account of the last three systems follows. 

352. Polar Coordinates. Consider the line OP drawn from 
the origin to any point P (Fig. 277). Let a, y8, y be the 
direction angles of OP, called the radius vector, and let p be the 
length of the radius vector. The four quantities a, p, y, p are 
called the polar coordinates of P. 

Conversely, any four quantities a, fi, y, p, with the restric- 
tion that cos2 a 4- cos^ ^ -)- cos^ y = 1, determine a point whose 
polar coordinates are a, p, y, p. 

Prove that tlie equations of transformation from rectangular to polar 
coordinates are. 



(10) 



x = p cos a, y = p cos p, z = p cos 7, p2 = x^^ +y^ -h z^ 



353. Spherical Coordinates. Any point P in space de- 
termines (Fig. 278) the radius vector OP(=p), the angle <f> 
between the radius vector and the 2!-axis, and 
the angle 6 between the a>-axis and the pro- 
jection of the radius vector on the a;2^-plane. 
The quantities p, 0, <f> are called the spherical 
coordinates of the point P. The angle <^ is 
known as the colatitude, and the angle 6 as 
the longitude. 

Conversely, any three quantities p, 6, <f> determine in space a 
point P whose spherical coordinates are p, 6, <i>. 

Prove that the equations of transformation from rectangular to spheri- 
cal coordinates are, 
(11) X = pBind coB(p, y = pain6sm(f>, z — p cos 6. 




Fia. 278 




XXII, § 354] QUADRIC SURFACES 531 

354. Cylindrical Coordinates. Any point P in space de- 
termines (Fig. 279) its distance z from the ^p 
ajy-plane and the polar coordinates r, 6 of 
the point P' which is the projection of P on 
the a;?/-plane. These three quantities r, 0, z 
are called the cylindrical coordinates of P. 
Conversely, any three quantities r, B, z deter- 
mine a point whose cylindrical coordinates they are. 

Prove that the equations of transformations from rectangular to cylin- 
drical coordinates are, 
(12) x = rQ.os,e, ?/ = rsin^, z = z. 

EXERCISES 

1. Express each of the following loci in spherical coordinates. 

(a) a;2 -I- ?/2 + z^ = 9. (6) a;2 _|. ^a _ 4^2 _ 0. (c) 4x2 + Qy'2 - z^ - 36. 

2. Express each of the following loci in polar coordinates. 

(a) ic2 + 2/2 4. 2;2 = 16. (6) X + 1/ = 0. (c) 2 x^ - y'^ - ^^ = 0. 

3. Express each of the following loci in cylindrical coordinates, 
(a) x^ + y'^ = 9. (6) x^ + y^ + z^ = 9. (c) z'^ - x'^ -\- y^ = 6. 

4. Express the distance between two points in polar coordinates. 

5. Find the polar, spherical, and cylindrical coordinates of the points 
whose rectangular coordinates are (2, 1, 4), (3, 3, 3). 

6. What is the locus of points for which 

(a) ^ = a constant, = a constant (spherical coordinates) ? 
(6) r = a constant, 6 z= sl constant (cylindrical coordinates) ? 

7. Find the general equation of a plane in polar coordinates. 

8. Find the general equation of a plane in spherical coordinates; 
in cylindrical coordinates. 

9. Show that in polar co5rdinates a point may be regarded as the 
intersection of a sphere and three cones of revolution which have an 
element in common. 

10. Show that in spherical coordinates a point may be regarded as 
the intersection of a sphere, a plane, and a cone of revolution which are 
mutually perpendicular. 

11. The spherical coordinates of a point are 5, 7r/4, 7r/6; find its 
rectangular coordinates ; its polar coordinates; its cylindrical coordinates. 



TABLES 

TO 

FOUE DECIMAL PLACES 



534 



Powers and Roots 



Squares and Cubes Square Roots and Cube Roots 



No. 


Sqitaee 


OUBK 


Square 
Root 


Cube 
Root 


No. 


Squabe 


Cttbe 


Square 
Root 


Cube 
Root 


1 


1 


1 


1.000 


1.000 


61 


2,601 


132,651 


7.141 


3.708 


2 


4 


8 


1.414 


1.260 


62 


2,704 


140,608 


7.211 


3.733 


3 


9 


27 


1.732 


1.442 


63 


2,809 


148,877 


7.280 


3.756 


4 


16 


64 


2.000 


1.587 


64 


2,916 


157,464 


7.348 


3.780 


5 


25 


125 


2.236 


1.710 


55 


3,025 


l(i6,375 


7.416 


3.803 


6 


36 


216 


2.449 


1.817 


56 


3,136 


175,616 


7.483 


3.826 


7 


49 


343 


2.646 


1.913 


57 


3,249 


185,193 


7.550 


3.849 


8 


64 


512 


2.828 


2.000 


58 


3,364 


195,112 


7.616 


3 871 


9 


81 


729 


3.000 


2.080 


59 


3,481 


205,379 


7.681 


3.893 


10 


100 


1,000 


3.162 


2.154 


60 


3,600 


216,000 


7.746 


3.915 


11 


121 


1,331 


3.317 


2.224 


61 


3,721 


226,981 


7.810 


3.936 


12 


144 


1,728 


3.464 


2.289 


62 


3,844 


238,328 


7.874 


3.958 


13 


169 


2,197 


3.606 


2.351 


63 


3,969 


250,047 


7.937 


3.979 


14 


196 


2,744 


3.742 


2.410 


64 


4,096 


262,144 


8.000 


4.000 


15 


225 


3,375 


3.873 


2.466 


65 


4,225 


274,625 


8.062 


4.021 


16 


256 


4,096 


4.000 


2.520 


66 


4,356 


287,496 


8.124 


4.041 


17 


289 


4,913 


4.123 


2.571 


67 


4,489 


300,763 


8.185 


4.062 


18 


324 


5,832 


4.243 


2.621 


68 


4,624 


314,432 


8.246 


4.082 


19 


361 


6,859 


4.359 


2.668 


69 


4,761 


328,509 


8.307 


4.102 


20 


400 


8,000 


4.472 


2.714 


70 


4,900 


343,000 


8.367 


4.121 


21 


441 


9,261 


4.583 


2.759 


71 


5,041 


357,911 


8.426 


4.141 


22 


484 


10,648 


4.690 


2.802 


72 


5,184 


373,248 


8.485 


4.160 


23 


529 


12,167 


4.796 


2.844 


73 


5,329 


389,017 


8.544 


4.179 


24 


576 


13,824 


4.899 


2.884 


74 


5,476 


405,224 


8.602 


4.198 


25 


625 


15,625 


5.000 


2.924 


75 


5,625 


421,875 


8.660 


4.217 


26 


676 


17,576 


5.099 


2.962 


76 


5,776 


438,976 


8.718 


4.236 


27 


729 


19,683 


5.196 


3.000 


77 


5,929 


456,533 


8.775 


4.254 


28 


784 


21,952 


5.292 


3.037 


78 


6,084 


474,552 


8.832 


4.273 


29 


841 


24,389 


5.385 


3.072 


79 


6,241 


493,039 


8.888 


4.291 


30 


900 


27,000 


6.477 


3.107 


80 


6,400 


512,000 


8.944 


4.309 


31 


961 


29,791 


5.568 


3.141 


81 


6,561 


531,441 


9.000 


4.327 


32 


1,024 


32,768 


5.657 


3.175 


82 


6,724 


551,368 


9.055 


4.344 


33 


1,089 


35,937 


5.745 


3.208 


83 


6,889 


571,787 


9.110 


4.362 


34 


1,156 


39,304 


5.831 


3.240 


84 


7,056 


592,704 


9.165 


4.380 


35 


1,225 


42,875 


5.916 


3.271 


85 


7,225 


614,125 


9.220 


4.397 


36 


1,296 


46,656 


6.000 


3.302 


86 


7,396 


636,056 


9.274 


4.414 


37 


1,369 


50,653 


6.083 


3.332 


87 


7,569 


658,503 


9.327 


4.431 


38 


1,444 


54,872 


6.164 


3.362 


88 


7,744 


681,472 


9.381 


4.448 


39 


1,521 


59,319 


6.245 


3.391 


89 


7,921 


704,969 


9.434 


4.465 


40 


1,600 


64,000 


6.325 


3.420 


90 


8,100 


729,000 


9.487 


4.481 


41 


1,681 


68,921 


6.403 


3.448 


91 


8,281 


753,571 


9.539 


4.498 


42 


1,764 


74,088 


6.481 


3.476 


92 


8,464 


778,688 


9.592 


4.514 


43 


1,849 


79,507 


6.557 


3.503 


93 


8,649 


804,357 


9.644 


4.531 


44 


1,936 


85,184 


6.633 


3.530 


94 


8,836 


830,584 


9.695 


4.547 


45 


2,025 


91,125 


6.708 


3.557 


95 


9,025 


857,375 


9.747 


4.563 


46 


2,116 


97,336 


6.782 


3.583 


96 


9,216 


884,736 


9.798 


4.579 


47 


2,209 


103,823 


6.856 3.609 


97 


9,409 


912,673 


9.849 


4.595 


48 


2,304 


110,592 


6.928 3.634 


98 


9,604 


941,192 


9.899 


4.610 


49 


2,401 


117,649 


7.000 3.659 


99 


9,801 


970,299 


9.950 


4.626 


60 


2,500 


125,000 


7.071 3.684 


100 


10,000 


1,000,000 


10.000 


4.642 



For a more complete table, see The Macmillan Tables, pp. 94-111. 



Important Constants 

Certain Convenient Values for n = 1 to n 



535 



10 



71 


1/n 


Vn 


V^ 


n ! 


l/n\ 


LOGlO V 


1 


1.000000 


1.00000 


1.00000 


1 


1.0000000 


0.000000000 


2 


0.600000 


1.41421 


1.25992 


2 


0.5000000 


0.301029996 


3 


0.333333 


1.73205 


1.44225 





0.1666667 


0.477121255 


4 


0.250000 


2.00000 


1.58740 


24 


0.0416667 


0.602059991 


5 


0.200000 


2.23607 


1.70998 


120 


0.0083333 


0.698970004 


6 


0.166667 


2.44949 


1.81712 


720 


0.0013889 


0.778151250 


7 


0.142857 


2.64575 


1.91293 


5040 


0.0001984 


0.845098040 


8 


0.125000 


2.82843 


2.00000 


40320 


0.0000248 


0.903089987 


9 


0.111111 


3.00000 


2.08008 


362880 


0.0000028 


0.954242509 


10 


0.100000 


3.16228 


2.15443 


3628800 


0.0000003 


1.000000000 



Logarithms of Important Constants 



n =>= NUiMBBH 


A^ALUE OF n 


LoGio n 


TT 


3.14159265 


0.49714987 


l-i-TT 


0.31830989 


9.50285013 


7r2 


9.86960440 


0.99429975 


V^ 


1.77245385 


0.24857494 


e = Napierian Base 


2.71828183 


0.43429448 


M= logio e 


0.43429448 


9.63778431 


l-4-i»/=logelO 


2.30258509 


0.36221569 


180 -J- TT = degrees in 1 radian 


57.2957795 


1.75812262 


TT -M80 = radians in 1° 


0.01745329 


8.24187738 


TT -4- 10800 = radians in 1' 


0.0002908882 


6.4(i372613 


TT -T- 648000 = radians in 1" 


0.000004848136811095 


4.68557487 


sin 1" 


0.000004848136811076 


4.68557487 


tan 1" 


0.000004848136811152 


4.68557487 


centimeters in 1 ft. 


30.480 


1.4840158 


feet in 1 cm. 


0.032808 


8.5159842 


inches in 1 m. 


39.37 (exact legal value) 


1.5951654 


pounds in 1 kg. 


2.20462 


0.3433340 


kilograms in 1 lb. 


0.453593 


9.65666(50 


g (average value) 


32.16 ft./sec./sec. 


1.5073 




= 981 cm./sec./sec. 


2.9916690 


-weight of 1 cu. ft. of water 


62.425 lb. (max. density) 


1.7953586 


weight of 1 cu. ft. of air 


0.0807 lb. (at 32° F.) 


8.907 


cu. in. in 1 (U. S.) gallon 


231 (exact legal value) 


2.3636120 


ft. lb. per sec. in 1 H. P. 


550. (exact legal value) 


2.7403627 


kg. m. per sec. in 1 H. P. 


76.0404 


1.8810445 


watts in 1 H. P. 


745.957 


2.8727135 



536 


) 








Four Place Logarithms 






N 





I 


2 


3 


4 


5 


6 


7 


8 


9 


12 3 


4 5 6 


7 8 9 


10 


0000 


0043 


0080 


0128 


0170 


0212 


0253 


02f)4 


0334 


0374 


4 812 


17 21 25 


29 33 37 


11 
12 
13 

14 
15 

IG 

17 

18 
19 


0414 
0702 
1139 

1461 
1701 
2041 

2304 
2553 

2788 


0453 

0328 
1173 

1492 

1790 
2068 

2330 

2577 
2810 


0492 
0864 
1206 

1523 
1818 
2095 

2355 

2.'>01 
2833 


0531 
0899 
1239 

1553 

1847 
2122 

2380 
2625 
2856 


0569 
0934 
1271 

1584 
1875 
2148 

2405 

2648 
2878 


0607 
0969 
1303 

1614 

1903 
2175 

2430 
2672 
2900 


0645 
1004 
1335 

1644 
1931 
2201 

245". 
2695 
2923 


0682 
1038 
1367 

1673 
1959 

2227 

2480 

2718 
2015 


0719 
1072 
1399 

1703 
1987 
2253 

2504 
2742 
2967 


0755 
1106 
1430 

1732 
2014 
2279 

2529 
2765 
2989 


4 8 11 
3 7 10 
3 6 10 

3 6 9 
3 6 8 
3 6 8 

2 6 7 

2 5 7 
2 4 7 


15 If, 23 
14 17 21 
13 16 19 

12 16 18 
11 14 17 
11 13 16 

10 12 16 
9 1214 
9 11 13 


26 30 34 
24 28 31 
23 26 29 

21 24 27 
20 22 25 
18 21 24 

17 20 22 
16 19 21 
16 18 20 


20 

21 
22 
23 

21 
25 

26 

27 
28 
29 

30 

31 
32 
33 

34 
35 

36 

37 

38 
39 


3010 


3032 


3054 


3075 


3096 


3118 


3139 


3160 


3181 


3201 


2 4 6 


8 1113 


16 17 19 


3222 
3424 
3617 

3802 
3979 
4150 

4314 
4472 
4624 

4771 

4914 
5051 
5185 

5315 
5441 
5563 

5682 
5798 
5911 


3213 
3444 
3636 

3820 
3997 
4166 

4330 

4487 
4639 

4786 

4928 
5065 
5198 

5328 
5453 
5575 

5694 

5809 
5922 


3263 
34G4 
3655 

3838 
4014 
4183 

4346 

4502 
4651 

4800 

4942 
5079 
5211 

5340 
5465 
5587 

5705 

5821 
5933 


3284 
3483 
3674 

3856 
4031 
4200 

4362 
4518 
4(569 


3304 
3502 
3692 

3874 
4048 
4216 

4378 
4533 
4683 


3324 
3522 
3711 

3892 
4065 
4232 

4393 
4548 
4698 


3345 
3541 
3729 

3909 
4082 
4249 

4409 
4564 
4713 


3365 
3560 
3747 

3927 
4099 
4265 

4425 

4579 

4728 


3385 
3579 
3766 

3946 
4116 
4281 

4440 
4594 
4742 


3404 
3598 
3784 

3962 
4133 
4298 

4456 
4609 
4757 


2 4 6 
2 4 6 
2 4 6 

2 4 5 
2 4 6 
2 3 6 

2 3 5 
2 3 6 
13 4 


8 10 12 
8 10 12 
7 9 11 

7 9 11 
7 9 10 
7 8 10 

6 8 9 
6 8 9 
6 7 9 


14 16 18 
14 16 17 
13 16 17 

12 14 16 
12 14 16 
11 13 16 

11 12 14 
11 12 14 
10 12 13 


4814 

4955 
5092 
5224 

5353 

5478 
5599 

5717 

6832 
5944 


4829 


4843 


4857 


4871 


4886 


4900 


1 3 4 


6 7 9 


10 11 13 


4969 
5105 
5237 

5366 
5490 
5611 

5729 
5843 
6955 


4983 
5119 
5250 

5378 
5502 
6623 

5740 
5855 
5966 


4997 
5132 
6263 

5391 
5514 
6635 

6752 

C866 
6977 


5011 
6145 
5276 

6403 
5527 
6647 

6763 

5877 
5988 


6024 
6159 
6289 

5416 
5539 
6658 

5775 

6888 
5999 


5038 
5172 
6302 

6428 
6551 
6670 

6786 
5899 
6010 


1 3 4 
1 3 4 
13 4 

1 2 4 
12 4 
12 4 

12 4 
1 2 3 
1 2 3 


6 7 8 
5 7 8 

5 7 8 

6 6 8 
6 6 7 
5 6 7 

5 6 7 

6 6 7 
4 6 7 


10 11 12 
911 12 
9 1112 

910 11 
910 11 
8 1011 

8 9 11 
8 9 10 
8 910 


40 


6021 


6031 


6042 


6053 


6064 


6075 


6085 


609(5 


6107 


6117 


12 3 


4 6 6 


8 9 10 


41 

42 
43 

44 
45 
46 

47 
48 
49 


6128 
6232 
6335 

6435 

05 32 
G628 

(5721 
6812 
6902 


6138 
6213 
6345 

6444 
6542 
6637 

6730 
6821 
6911 


6149 
6253 
6355 

6454 
6551 
664(5 

6739 
6830 
6020 


6160 
6263 
6365 

6464 
65(51 
6656 

6749 
6839 
6928 


6170 
6274 
6375 

6474 
6571 
6665 

6758 
6848 
6937 


6180 
6284 
6385 

6484 
6580 
6675 

6767 
6857 
6946 


6191 
6294 
6395 

6493 

6590 
(5684 

6776 

6866 
6955 


6201 
6304 
6405 

6503 
6599 
6693 

6785 
6875 
69(54 


6212 
6314 
6416 

6513 
6609 
6702 

6794 
6884 
6972 


6222 
6325 
6425 

6622 
6618 
6712 

6803 
6893 
6981 


1 2 3 
12 3 
12 3 

1 2 3 
1 2 3 
1 2 3 

12 3 
1 2 3 
1 2 3 


4 5 6 
4 5 6 
4 5 6 

4 5 6 
4 6 6 
4 5 6 

4 5 6 
4 6 6 
4 4 6 


7 8 9 
7 8 9 
7 8 9 

7 8 9 
7 8 9 
7 7 8 

7 7 8 
7 7 8 
6 7 8 


50 


(5990 


6998 


7007 


7016 


7024 


7033 


7042 


7050 


7059 


7067 


12 3 


3 4 6 


6 7 8 


51 

52 
53 

54 


7076 
71G0 
7243 

7324 


7084 
7168 
7251 

7332 


7093 
7177 
7259 

7340 


7101 
7185 
7267 

7348 


7110 
7193 

7275 

7356 


7118 
7202 
7284 

7364 


7126 
7210 
7292 

7372 


7135 
7218 
7300 

7380 


7143 
7226 
7308 

7388 


7152 
7235 
7316 

739(5 


1 2 3 
1 2 3 
12 2 

1 2 2 


3 4 5 
3 4 5 
3 4 5 

3 4 6 


6 7 8 
6 7 7 
6 6 7 

6 6 7 


N 





1 


2 


3 


4 


5 


6 


7 


8 


9 


12 2 


4 5 6 


7 8 9 



The proportional parts are stated in full for every tenth at the right-hand side. 
The logarithm of any number of four significant figures can be read directly by add- 




ing the proportional part correspoTiding to the fourth figure to the tabular number 
correspondmg to the first three figures. There may be an error of 1 in the last place. 



538 



Four Place Trigonometric Functions 



[Characteristics of Logarith 


ms orni 


tted — 


determine by the usual rule from the value] 


Radiaks 


DT^aPTTfl 


Sine 


Tangent 


Cotangent 


Cosine 








i-.'£*ijri&x>i:<o 


Value 


Logio 


Value 


Logio 


Value 


Logio 


Value 


Logio 






.0000 
.0029 


0°00' 

10 


.0000 
.0029 




.0000 
.0029 








1.0000 


.0000 


90° 00' 


1.5708 
1.5679 


.4637 


.4637 


343.77 


.6363 


1.0000 


.0000 


50 


.0058 


20 


.0058 


.7()48 


.0058 


.7648 


171.89 


.2352 


1.0000 


.0000 


40 


1.5650 


.0087 


30 


.0087 


.9408 


.0087 


.9409 


114.59 


.0591 


1.0000 


.0000 


30 


1.5621 


.0116 


40 


.0116 


.0658 


.0116 


.0658 


85.940 


.9342 


.9999 


.0000 


20 


1.5592 


.0145 


50 


.0145 


.1627 


.0145 


.1627 


68.750 


.8373 


.9999 


.0000 


10 


1.5563 


.0175 


1°00' 


.0175 


.2419 


.0175 


.2419 


57.290 


.7581 


.9998 


.9999 


89° 00' 


1.5533 


.0204 


10 


.0204 


.3088 


.0204 


.3089 


49.104 


.6911 


.9998 


.9999 


50 


1.5504 


.0233 


20 


.0233 


.3668 


.0233 


.3669 


42.964 


.6331 


.9997 


.9999 


40 


1.547.-. 


.0202 


30 


.0262 


.4179 


.02()2 


.4181 


38.188 


.5819 


.99f)7 


.9999 


30 


1.5440 


.0291 


40 


.0291 


.4637 


.0291 


.4638 


34.368 


.5362 


.9996 


.9998 


20 


1.5417 


.0320 


50 


.0320 


.5050 


.0320 


.5053 


31.242 


.4947 


.9995 


.9998 


10 


1.5388 


.0349 


2° 00' 


.0349 


.5428 


.0349 


.5431 


28.636 


.4569 


.9994 


.99f)7 


88° 00' 


1.5359 


.0.378 


10 


.0378 


.5776 


.0378 


.5779 


26.432 


.4221 


.9993 


.9997 


50 


1.5330 


.0407 


20 


.0407 


.6097 


.0407 


.6101 


24.542 


.3899 


.9992 


.9996 


40 


1.5301 


.0436 


30 


.0436 


.6397 


.0437 


.6401 


22.904 


.3599 


.99{)0 


.9996 


30 


1.5272 


.0465 


40 


.0465 


.6677 


.0466 


.6682 


21.470 


.3318 


.9989 


.9995 


20 


1.5243 


.0495 


50 


.0494 


.6940 


.0495 


.6945 


20.20(5 


.3055 


.t)988 


.9995 


10 


1.5213 


.0524 


3° 00' 


.0523 


.7188 


.0524 


.7194 


19.081 


.2806 


.9986 


.9994 


87° 00' 


1.5184 


.0553 


10 


.0552 


.7423 


.0553 


.7429 


18.075 


.2571 


.9985 


.9993 


50 


1.5155 


.0582 


20 


.0581 


.7645 


.0582 


.7652 


17.169 


.2348 


.9983 


.9993 


40 


1.5126 


.0611 


30 


.0610 


.7857 


.0612 


.7865 


16..350 


.2135 


.9981 


.9992 


30 


1.5097 


.0640 


40 


.0640 


.8059 


.0641 


.8067 


15.605 


.1933 


.9980 


.9991 


20 


1.5068 


.0669 


50 


.0669 


.8251 


.0070 


.8261 


14.924 


.1739 


.9978 


.9990 


10 


1.5039 


.0698 


4° 00' 


.0698 


.8436 


.0699 


.8446 


14.301 


.1554 


.9976 


.9989 


86° 00' 


1.5010 


.0727 


10 


.0727 


.8613 


.0729 


.8624 


13.727 


.1376 


.9974 


.9989 


50 


1.4981 


.0756 


20 


.0756 


.8783 


.0758 


.8795 


13.197 


.1205 


.9971 


.9988 


40 


1.4952 


.0785 


30 


.0785 


.8946 


.0787 


.8960 


12.706 


.1040 


.9969 


.9987 


30 


1.4923 


.0814 


40 


.0814 


.9104 


.0816 


.9118 


12.251 


.0882 


.9967 


.9986 


20 


1.4893 


.0844 


50 


.0843 


.9256 


.0846 


.9272 


11.826 


.0728 


.9964 


.9985 


10 


1.4864 


.0873 


6° 00' 


.0872 


.9403 


.0875 


.9420 


11.430 


.0580 


.9962 


.9983 


86° 00' 


1.4835 


.0902 


10 


.0901 


.9545 


.0904 


.9563 


11.059 


.0437 


.9959 


.9982 


60 


1.4806 


.0931 


20 


.0929 


.9682 


.0934 


.9701 


10.712 


.0299 


.9957 


.9981 


40 


1.4777 


.0960 


30 


.0958 


.9816 


.0963 


.9836 


10.385 


.0164 


.9954 


.9980 


30 


1.4748 


.0985) 


40 


.0987 


.9945 


.0992 


.9966 


10.078 


.0034 


.9951 


.9979 


20 


1.4719 


.1018 


50 


.1016 


.0070 


.1022 


.0093 


9.7882 


.9907 


.9948 


.9977 


10 


1.4690 


.1047 


6° 00' 


.1045 


.0192 


.1051 


.0216 


9.5144 


.9784 


.9945 


.9976 


84° 00' 


1.4661 


.1076 


10 


.1074 


.0311 


.1080 


.0336 


9.2553 


.9664 


.9942 


.9975 


50 


1.4632 


.1105 


20 


.1103 


.0426 


.1110 


.0453 


9.0098 


.9547 


.9939 


.9973 


40 


1.4603 


.1134 


30 


.1132 


.0539 


.1139 


.0567 


8.7769 


.9433 


.993(5 


.9972 


30 


1.4573 


.1164 


40 


.1161 


.0648 


.1169 


.0678 


8.5555 


.9322 


.<)i)32 


.9971 


20 


1.4544 


.1193 


50 


.1190 


.0755 


.1198 


.0786 


8.3450 


.9214 


.9929 


.9909 


10 


1.4615 


.1222 


7° 00' 


.1219 


.0859 


.1228 


.0891 


8.1443 


.9109 


.9925 


.9968 


83° 00' 


1.4486 


.1251 


10 


.1248 


.0961 


.1257 


.0995 


7.9530 


.JK)05 


M^2 


.9966 


60 


1.4457 


.1280 


20 


.1276 


.1060 


.1287 


.109(i 


7.7704 


.8904 


.9918 


.9964 


40 


1.4428 


.1309 


30 


.1305 


.1157 


.1317 


.1194 


7.5958 


.880(i 


.9914 


.9^)63 


30 


1.4399 


.1338 


40 


.1334 


.1252 


.1346 


.1291 


7.4287 


.8709 


.9911 


.9961 


20 


1.4370 


.1367 


50 


.1363 


.1345 


.1376 


.1385 


7.2(i87 


.8615 


.9907 


.9959 


10 


1.4341 


.1396 


8° 00' 


.1392 


.14.36 


.1405 


.1478 


7.1154 


.8522 


.9903 


.9958 


82° 00' 


1.4312 


.1425 


10 


.1421 


.1525 


.1435 


.1.569 


6.9682 


.8431 


.9899 


.9<)56 


50 


1.4283 


.1454 


20 


.1449 


.1612 


.1465 


.1658 


6.8269 


.8342 


.9894 


.9954 


40 


1.4254 


.148'1 


30 


.1478 


.1697 


1495 


.1745 


6.6912 


.8255 


.9890 


.9952 


30 


1.4224 


.1513 


40 


.1507 


.1781 .1524 


.18.31 


().5606 


.8169 


.9886 


.9950 


20 


1.41^)5 


.1542 


50 


.1536 


.1863 .1.5.54 


.1915 


6.4348 


.8085 


.9881 


.9948 


10 


1.4166 


.1571 


9° 00' 


.1564 


.1943 


.1584 


.1997 


6.3138 


.8003 


.9877 


.9946 


81° 00' 


1.4137 






V^alne 


Logjo 


Value 


Lopin 


Value 


Lopio 


Value 


Logio 


Degrees 


Rapians 






Cosine 


Cotangent 


Tangent 


Sink 







Four Place Trigonometric Functions 



539 



[Charactoristics of Logarithms omitted — 


letermine by the usual rule from the value] 


Radians 


Degeees 


Sine 


Tangent 


Cotangent 


Cosine 










7alue 


Logio 


Value 


Logio 


Value 


Logio 


V^alue 


Logio 






.1571 


9° 00' 


.1564 


.1943 


.1584 


.1997 


6.3138 


.8003 


.9877 


.9946 


81° 00' 


1.4137 


.1600 


10 


.1593 


.2022 


.1614 


.2078 


6.1970 


.7922 


.9872 


.9944 


50 


1.4108 


.1029 


20 


.1622 


.2100 


.1044 


.2158 


6.0844 


.7842 


.98(38 


.9942 


40 


1.4079 


.1058 


30 


.1650 


.2176 


.1073 


.2236 


5.9758 


.7764 


.9863 


.9940 


30 


1.4050 


.1087 


40 


.1079 


.2251 


.1703 


.2313 


5.8708 


.7687 


.9858 


.9938 


20 


1.4021 


.1716 


50 


.1708 


.2324 


.1733 


.2389 


5.7694 


.7611 


.9853 


.9936 


10 


1.3992 


.1745 


10° 00' 


.1736 


.2397 


.1763 


.2463 


5.6713 


.7537 


.9848 


.9934 


80° 00' 


1.3963 


.1774 


10 


.1765 


.2468 


.1793 


.2536 


5.5764 


.7464 


.9843 


.9931 


50 


1.3934 


.1804 


20 


.1794 


.2538 


.1823 


.2609 


5.4845 


.7391 


.9838 


.9929 


40 


1.3904 


.1833 


30 


.1822 


.2600 


.1853 


.2680 


5.3955 


.7320 


.9833 


.992; 


30 


1.3875 


.1802 


40 


.1851 


.2674 


.1883 


.2750 


5.3093 


.7250 


.9827 


.9924 


20 


1.3840 


.1891 


50 


.1880 


.2740 


.1914 


.2819 


5.2257 


.7181 


.9822 


.9922 


10 


1.3817 


.1920 


11°00' 


.1908 


.2806 


.1944 


.2887 


5.1446 


.7113 


.9816 


.9919 


79° 00' 


1.3788 


.1949 


10 


.1937 


.2870 


.1974 


.2953 


5.0058 


.7047 


.9811 


.9917 


50 


1.3759 


.1978 


20 


.1965 


.2934 


.2004 


.3020 


4.9894 


.6980 


.9805 


.9914 


40 


1.3730 


.2007 


30 


.1994 


.25)97 


.2035 


.3085 


4.9152 


.6915 


.9799 


.9912 


30 


1.3701 


.2036 


40 


.2022 


.3058 


.2065 


.3149 


4.8430 


.6851 


.9793 


.9909 


20 


1.3672 


.2065 


50 


.2051 


.3119 


.2095 


.3212 


4.7729 


.6788 


.9787 


.9907 


10 


1.3&43 


.2094 


12° 00' 


.2079 


.3179 


.2126 


.3275 


4.7046 


.6725 


.9781 


.9904 


78° 00' 


1.3014 


.2123 


10 


.2108 


.3238 


.2156 


.3330 


4.6382 


.6664 


.9775 


.9901 


■50 


1.3584 


.2153 


20 


.2136 


.3296 


.2186 


.3397 


4.5736 


.6603 


.9769 


.9899 


40 


1.3555 


.2182 


30 


.2164 


.3353 


.2217 


.3458 


4.5107 


.6542 


.9763 


.9896 


30 


1.3526 


.2211 


40 


.2193 


.3410 


.2247 


.3517 


4.4494 


.6483 


.9757 


.9893 


20 


1.3497 


.2240 


60 


.2221 


.3466 


.2278 


.3576 


4.3897 


.6424 


.9750 


.9890 


10 


1.3468 


.2269 


13° 00' 


.2250 


.3521 


.2309 


.3634 


4.3315 


.636() 


.9744 


.9887 


77° 00' 


1.3439 


.2298 


10 


.2278 


.3575 


.2339 


.3691 


4.2747 


.6309 


.9737 


.9884 


50 


1.3410 


.2327 


20 


.2306 


.3629 


.2370 


.3748 


4.2193 


.0252 


.9730 


.9881 


40 


1.3381 


.2356 


30 


.2334 


.3682 


.2401 


.3804 


4.1653 


.6196 


.9724 


.9878 


30 


1.3352 


.2385 


40 


.2363 


.3734 


.2432 


.3859 


4.1126 


.6141 


.9717 


.9875 


20 


1.3323 


.2414 


60 


.2391 


.3786 


.2462 


.3914 


4.0611 


.6080 


.9710 


.9872 


10 


1.3294 


.2443 


14° 00' 


.2419 


.3837 


.2493 


.3968 


4.0108 


.6032 


.9703 


.9869 


76° 00' 


1.3265 


.2473 


10 


.2447 


.3887 


.2524 


.4021 


3.9017 


.5979 


.96% 


.9866 


50 


1.3235 


.2502 


20 


.2476 


.3937 


.2555 


.4074 


3.9136 


.5926 


.9689 


.9863 


40 


1.3206 


.2531 


30 


.2504 


.3986 


.2586 


.4127 


3.8667 


.5873 


.9681 


.9859 


30 


1.3177 


.2560 


40 


.2532 


.4035 


.2617 


.4178 


3.8208 


.5822 


.9674 


.9856 


20 


1.3148 


.2589 


50 


.2560 


.4083 


.2648 


.4230 


3.7760 


.5770 


.9(367 


.9853 


10 


1.3119 


.2618 


15°00' 


.2588 


.4130 


.2679 


.4281 


3.7321 


.5719 


.9659 


.9849 


75° 00' 


1.3090 


.2647 


10 


.2616 


.4177 


.2711 


.4331 


3.0891 


.5669 


.9652 


.9846 


50 


1.3061 


.2676 


20 


.2644 


.4223 


.2742 


.4381 


3.6470 


.5619 


.9644 


.9843 


40 


1.3032 


.2705 


30 


.2672 


.4269 


.2773 


.4430 


3.6059 


.5570 


.9(>36 


.9839 


30 


1.3003 


.2734 


40 


.2700 


.4314 


.2805 


.4479 


3.5656 


.5521 


.9(528 


.9836 


20 


1.2974 


.2763 


60 


.2728 


.4359 


.2836 


.4527 


3.5261 


.5473 


.9621 


.9832 


10 


1.2945 


.2793 


16° 00' 


.2756 


.4403 


.2867 


.4575 


3.4874 


.5425 


.9613 


.9828 


74° 00' 


1.2915 


.2822 


10 


.2784 


.4447 


.2899 


.4622 


3.4495 


.5378 


.9605 


.9825 


50 


1.2886 


.2851 


20 


.2812 


.4491 


.2931 


.4669 


3.4124 


.5331 


.9596 


.9821 


40 


1.2857 


.2880 


30 


.2840 


.4533 


.2962 


.4716 


3.3759 


.5284 


.9588 


.9817 


30 


1.2828 


.2909 


40 


.2868 


.4576 


.2994 


.4762 


3.3402 


.5238 


.9580 


.9814 


20 


1.2799 


.2938 


50 


.2896 


.4618 


.3026 


.4808 


3.3052 


.5192 


.9572 


.9810 


10 


1.2770 


.2967 


17° 00' 


.2924 


.4659 


.3057 


.4853 


3.2709 


.5147 


.9563 


.9806 


73° 00' 


1.2741 


.2996 


10 


.2952 


.4700 


.3089 


.4898 


3.2371 


.5102 


.9555 


.9802 


50 


1.2712 


.3025 


20 


.2979 


.4741 


.3121 


.4943 


3.2041 


.5057 


.9546 


.9798 


40 


1.2683 


.3054 


30 


.3007 


.4781 


.3153 


.4987 


3.1716 


.5013 


.9537 


.9794 


30 


1.2654 


.3083 


40 


.3035 


.4821 


.3185 


.5031 


3.1397 


.4969 


.9528 


.9790 


20 


1.2625 


.3113 


50 


.3062 


.4861 


.3217 


.5075 


3.1084 


.4925 


.9520 


.9786 


10 


1.2696 


.3142 


18^00' 


.3090 


.4900 


.3249 


.5118 


3.0777 


.4882 


.9511 


.9782 


72° 00' 


1.2566 






Value 


LoSio 


Value 


Loffio 


Value 


T^'^ffio 


Value 


Logjo 


Degeees 


Radians 






Co'JINE 


Cotangent 


Tangent 


Sine 







540 



Four Place Trigonometric Functions 



[Characteristics of Logarithms omitted — 


ietermine by the usual rule from the value] 


Kadians 


Deobees 


Sine 


Tangent 


Cotangent 


Cosine 










Value 


Logio 


Value 


Logio 


Value 


Logio 


Value 


Lopio 






.3142 


18° 00' 


.3090 


.4900 


.3249 


.5118 


3.0777 


.4882 


.9511 


.9782 


72° 00' 


1.2566 


.3171 


10 


.3118 


.4939 


.3281 


.6161 


3.0475 


.4839 


.9502 


.9778 


60 


1.2537 


.3200 


20 


.3145 


.4977 


.3314 


.6203 


3.0178 


.4797 


.9492 


.9774 


40 


1.2508 


.3229 


30 


.3173 


.5015 


.3346 


.6245 


2.9887 


.4756 


.9483 


.9770 


30 


1.2479 


.3258 


40 


.3201 


.5052 


.3378 


.6287 


2.9600 


.4713 


.9474 


.9765 


20 


1.2450 


.3287 


50 


.3228 


.5090 


.3411 


.5329 


2.9319 


.4671 


.9465 


.9761 


10 


1.2421 


.3316 


19° 00' 


.3256 


.5126 


.3443 


.5370 


2.9042 


.4630 


.9455 


.9767 


71° 00' 


1.2392 


.3346 


10 


.3283 


.5163 


.3476 


.5411 


2.8770 


.4689 


.9446 


.9752 


60 


1.2363 


.3374 


20 


.3311 


.5199 


.3508 


.5451 


2.8502 


.4649 


.9436 


.9748 


40 


1.2334 


.3403 


30 


.3338 


.5235 


.3541 


.5491 


2.8239 


.4509 


.9426 


.9743 


30 


1.2305 


.3432 


40 


.3365 


.5270 


.3574 


.6531 


2.7980 


.4469 


.9417 


.9739 


20 


1.2275 


.3462 


50 


.3393 


.5306 


.3607 


.5571 


2.7725 


.4429 


.9407 


.9734 


10 


1.2246 


.3491 


20° 00' 


.3420 


.5341 


.3640 


.6611 


2.7476 


.4389 


.9397 


.9730 


70° 00' 


1.2217 


.3520 


10 


.3448 


.5375 


.3673 


.5650 


2.7228 


.4350 


.9387 


.9725 


50 


1.2188 


.3549 


20 


.3475 


.5409 


.3706 


.5689 


2.6986 


.4311 


.9377 


.9721 


40 


1.2159 


.3578 


30 


.3502 


.5443 


.3739 


.6727 


2.6746 


.4273 


.9367 


.9716 


30 


1.2130 


.3607 


40 


.3529 


.5477 


.3772 


.6766 


2.6511 


.4234 


.9356 


.9711 


20 


1.2101 


.3636 


50 


.3557 


.5510 


.3805 


.5804 


2.6279 


.4196 


.9346 


.9706 


10 


1.2072 


.3665 


21° 00' 


.3584 


.5643 


.3839 


.5842 


2.6051 


.4158 


.9336 


,9702 


69° 00' 


1.2043 


.3694 


• 10 


.3611 


.5576 


.3872 


.6879 


2.5826 


.4121 


.9325 


.9697 


60 


1.2014 


.3723 


20 


.3638 


.5609 


.3906 


.5917 


2.5605 


.4083 


.9315 


.9692 


40 


1.1985 


.3752 


30 


.3m5 


.5641 


.3939 


.5954 


2.5386 


.4046 


.9304 


.9687 


30 


1.1956 


.3782 


40 


.3692 


.5673 


.3973 


.5991 


2.5172 


.4009 


.9293 


.9682 


20 


1.1926 


.3811 


50 


.3719 


.5704 


.4006 


.6028 


2.4960 


.3972 


.9283 


.9677 


10 


1.1897 


.3840 


22° 00' 


.3746 


.6736 


.4040 


.6064 


2.4751 


.3936 


.9272 


.9672 


68° 00' 


1.1868 


.3869 


10 


.3773 


.5767 


.4074 


.6100 


2.4545 


.3900 


.9261 


.9667 


50 


1.1839 


.3898 


20 


.3800 


.5798 


.4108 


.6136 


2.4342 


.3864 


.9250 


.9661 


40 


1.1810 


.3927 


30 


.3827 


.6828 


.4142 


.6172 


2.4142 


.3828 


.9239 


.9(556 


30 


1.1781 


.3956 


40 


.3854 


.5859 


.4176 


.6208 


2.3945 


.3792 


.9228 


.9651 


20 


1.1752 


.3985 


50 


.3881 


.5889 


.4210 


.6243 


2.3750 


.3767 


.9216 


.9646 


10 


1.1723 


.4014 


23° 00' 


.3907 


.6919 


.4245 


.6279 


2.3559 


.3721 


.9205 


.9640 


67° 00' 


1.1694 


.4043 


10 


.3934 


.5948 


.4279 


.6314 


2.3309 


.3686 


.9194 


.9635 


50 


1.1(>65 


.4072 


20 


.3961 


.5978 


.4314 


.6348 


2,3183 


.3652 


.9182 


.9629 


40 


1.1636 


.4102 


30 


.3987 


.6007 


.4348 


.6383 


2.2998 


.3617 


.9171 


.9624 


30 


1.160(5 


.4131 


40 


.4014 


.6036 


.4383 


.6417 


2.2817 


.3583 


.9159 


.9618 


20 


1.1577 


.4160 


60 


.4041 


.6065 


.4417 


.6462 


2.2637 


.3548 


.9147 


.9613 


10 


1.1548 


.4189 


24° 00' 


.4067 


.6093 


.4452 


.6486 


2.24(50 


.3514 


.9135 


.9607 


66° 00' 


1.1519 


.4218 


10 


.4094 


.6121 


.4487 


.6520 


2.2286 


.3480 


.9124 


.9602 


50 


1.1490 


.4247 


20 


.4120 


.6149 


.4522 


.(5553 


2.2113 


.3447 


.9112 


.9596 


40 


1.1461 


.4276 


30 


.4147 


.6177 


.4557 


.6587 


2.1943 


.3413 


.9100 


.9590 


30 


1.1432 


.4305 


40 


.4173 


.6206 


.4592 


.6620 


2.1775 


.3380 


.9088 


.9584 


20 


1.1403 


.4334 


50 


.4200 


.6232 


.4628 


.6654 


2.1609 


.3346 


.9075 


.9579 


10 


1.1374 


.4363 


25° 00' 


.4226 


.6259 


.4663 


.6687 


2.1445 


.3313 


.9063 


.9573 


65° 00' 


1.1345 


.4392 


10 


.4253 


.6286 


.4699 


.6720 


2.1283 


.3280 


.<X)51 


.95(57 


50 


1.1316 


.4422 


20 


.4279 


.6313 


.4734 


.6752 


2.1123 


.3248 


.9038 


.9561 


40 


1.128(5 


.4451 


30 


.4305 


.6340 


.4770 


.6785 


2.0965 


.3215 


.9026 


.95,^5 


30 


1.1257 


.4480 


40 


.4331 


.6.366 


.4806 


.6817 


2.0809 


.3183 


.9013 


.9549 


20 


1.1228 


.4509 


50 


.4358 


.6392 


.4841 


.6850 


2.0656 


.3150 


.9001 


.9543 


10 


1.1199 


.4538 


26° 00' 


.4384 


.6418 


.4877 


.6882 


2.0503 


.3118 


.8988 


.9537 


64° 00' 


1.1170 


.4567 


10 


.4410 


.6444 


.4913 


.6914 


2.0353 


.3086 


.8975 


.9.530 


50 


1.1141 


.4596 


20 


.4436 


.6470 


.4950 


.6946 


2.0204 


.3054 


.89(52 


.9624 


40 


1.1112 


.4625 


30 


.4462 


.0495 


.4986 


.(5977 


2.0057 


.3023 


.8949 


.9518 


30 


1.1083 


.4654 


40 


.4488 


.6621 


.5022 


.7009 


1.9^)12 


.2991 


.89:^5 


.9512 


20 


1.1054 


.4083 


60 


.4514 


.6646 


.5059 


.7040 


1.9768 


.2960 


.8923 


.9505 


10 


1.1025 


.4712 


27° 00' 


.4540 


.6670 


.6096 


.7072 


1.9626 


.2928 


.8910 


.9499 


63° 00' 


1.0996 






Value 


Logio 


Value 


Lopio 


Value 


Logio 


Value 


Log,o 


Degrees 


Radians 






Cosine 


Cotangent 


Tangent 


Sine 







Four Place Trigonometric Functions 541 

[Oharaoteristics of Logarithms omitted — determine by the usual rule from the value] 



Radians 


Dbgebes 


SI.B 


Tangent 


Cotangent 


Cosine 






Value 


Loffio 


Value 


Logio 


Value Logio 


Value 


Logio 






.4712 


27° 00' 


.4540 


.6570 


.5095 


.7072 


1.9626 .2923 


.8910 


.9499 


63° 00' 


1.0996 


.4741 


10 


.4566 


.6595 


.5132 


.7103 


1.9486 .2897 


.8897 


.9492 


50 


1.0966 


.4771 


20 


.4592 


.6620 


.5169 


.7134 


1.9347 .2866 


.8884 


.948(5 


40 


1.0937 


.4800 


30 


.4617 


.6644 


.5206 


.7165 


1.9210 .2835 


.8870 


.9479 


30 


1.0908 


.4829 


40 


.4643 


.6663 


.5243 


.7196 


1.9074 .2804 


.8857 


.9473 


20 


1.0879 


.4858 


60 


.4669 


.6692 


.5280 


.7226 


1.8940 .2774 


.8843 


.9466 


10 


1.0850 


.4887 


28° 00' 


.4695 


.6716 


.5317 


.7257 


1.8807 .2743 


.8829 


.9459 


62° 00' 


1.0821 


.4Q16 


10 


.4720 


.6740 


.5354 


.7287 


1.8676 .2713 


.8816 


.9453 


50 


1.0792 


.4945 


20 


.4746 


.6763 


.5392 


.7317 


1.8546 .2683 


.8802 


.9446 


40 


1.0703 


.4974 


30 


.4772 


.6787 


.5430 


.7348 


1.8418 .2652 


.8788 


.9439 


30 


1.0734 


.5003 


40 


.4797 


.6810 


.5167 


.7378 


1.8291 .2(522 


.8774 


.9432 


20 


1.0705 


.5032 


50 


.4823 


.6833 


.5505 


.7408 


1.8165 .2592 


.8760 


.9425 


10 


1.0676 


.5061 


29° 00' 


.4848 


.6856 


.5543 


.7438 


1.8040 .2562 


.8746 


.9418 


61° 00' 


1.0647 


.5091 


10 


.4874 


.6878 


.5581 


.7467 


1.7917 .2533 


.8732 


.9411 


50 


1.0617 


.5120 


20 


.4899 


.6901 


.5619 


.7497 


1.7796 .2503 


.8718 


.9404 


40 


1.0588 


.5149 


30 


.4924 


.6923 


.5658 


.7526 


1.7675 .2474 


.8704 


.9397 


30 


1.0559 


.5178 


40 


.4950 


.6946 


.5696 


.7556 


1.7556 .2444 


.8689 


.9390 


20 


1.0530 


.5207 


50 


.4975 


.6968 


.5735 


.7585 


1.7437 .2415 


.8675 


.9383 


10 


1.0501 


.5236 


30° 00' 


.5000 


.6990 


.5774 


.761i 


1.7321 .2386 


.8660 


.9375 


60° 00' 


1.0472 


.5265 


10 


.5025 


.7012 


.7(544 


1.7205 .2356 


.8646 


.9368 


50 


1.0443 


.5294 


20 


.50jO 


.7033 


.5851 


.7673 


1.7090 .2327 


.8631 


.9361 


40 


1.0414 


.5323 


30 


.5075 


.7055 


.5890 


.7701 


1.6977 .2299 


.8616 


.9353 


30 


1.0385 


.5352 


40 


.5100 


.7076 


.5930 


.7730 


1.6864 .2270 


.8601 


.9346 


20 


1.0356 


.5381 


50 


.5125 


.7097 


.5969 


.7759 


1.6753 .2241 


.8587 


.9338 


10 


1.0327 


.5411 


31°00' 


.5150 


.7118 


.6009 


.7788 


1.6643 .2212 


.8572 


.9331 


59° 00' 


1.0297 


.5440 


10 


.5175 


.7139 


.6048 


.7816 


1.6534 .2184 


.8557 


.9323 


.50 


1.0268 


.5469 


20 


.5200 


.7160 


.6088 


.7845 


1.6426 .2155 


.8542 


.9315 


40 


1.0239 


.5498 


30 


.5225 


.7181 


.6128 


.7873 


1.6319 .2127 


.8526 


.9308 


30 


1.0210 


.5527 


40 


.5250 


.7201 


.6168 


.7902 


1.6212 .2098 


.8511 


.9300 


20 


1.0181 


.5556 


50 


.5275 


.7222 


.6208 


.7930 


1.6107 .2070 


.8496 


.9292 


10 


1.0152 


.5585 


32° 00' 


.5299 


.7242 


.6249 


.7958 


1.6003 .2042 


.8480 


.9284 


58° 00' 


1.0123 


.5614 


10 


.5324 


.7262 


.6289 


.7986 


1.5900 .2014 


.8465 


.9276 


50 


1.0094 


.5643 


20 


.5348 


.7282 


.6330 


.8014 


1.5798 .1986 


.8450 


.9268 


40 


1.0065 


.5672 


30 


.5373 


.7302 


.6371 


.8042 


1.5697 .1958 


.8434 


.9260 


30 


1.0036 


.5701 


40 


.5398 


.7322 


.6412 


.8070 


1.5597 .1930 


.8418 


.9252 


20 


1.0007 


.5730 


50 


.5422 


.7342 


.6453 


.8097 


1.5497 .1903 


.8403 


.9244 


10 


.9977 


.5760 


33° 00' 


.5446 


.7361 


.6494 


.8125 


1.5399 .1875 


.8387 


.9236 


57° 00' 


.9948 


.5789 


10 


.5471 


.7380 


.6536 


.8153 


1.5301 .1847 


.8371 


.9228 


50 


.9919 


.5818 


20 


.5495 


.7400 


.6577 


.8180 


1.5204 .1820 


.8.355 


.9219 


40 


.9890 


.5847 


30 


.5519 


.7419 


.6619 


.8208 


1.5108 .1792 


.8339 


.9211 


30 


.9861 


.587(5 


40 


.5544 


.7438 


.6(561 


.8235 


1.5013 .1765 


.8323 


.9203 


20 


.9832 


.5905 


50 


.5568 


.7457 


.6703 


.8263 


1.4919 .1737 


.8307 


.9194 


10 


.9803 


.5934 


34° 00' 


.5592 


.7476 


.6745 


.8290 


1.4826 .1710 


.8290 


.9186 


66° 00' 


.9774 


.5903 


10 


.5616 


.7494 


.6787 


.8317 


1.4733 .1683 


.8274 


.9177 


50 


.9745 


.5992 


20 


.5640 


.7513 


.6830 


.8344 


1.4641 .1656 


.8258 


.9169 


40 


.9716 


.6021 


30 


.5664 


.7531 


.6873 


.8371 


1.4550 .1629 


.8241 


.9160 


30 


.9687 


.6050 


40 


.5688 


.7550 


.6916 


.8398 


1.4460 .1602 


.8225 


.9151 


20 


.9657 


.6080 


50 


.5712 


.7568 


.6959 


.8425 


1.4370 .1575 


.8208 


.9142 


10 


.9628 


.6109 


35° 00' 


.5736 


.7586 


.7002 


.8452 


1.4281 .1548 


.8192 


.9134 


55° 00' 


.9599 


.6138 


10 


.5760 


.7604 


.7046 


.8479 


1.4193 .1521 


.8175 


.9125 


50 


.9570 


.6167 


20 


.5783 


.7622 


.7089 


.8506 


1.4106 .1494 


.8158 


.9116 


40 


.9541 


.6196 


30 


.5807 


.7640 


.7133 


.8533 


1.4019 .1467 


.8141 


.9107 


30 


.9512 


.6225 


40 


.5831 


.7657 


.7177 


.8559 


1.3934 .1441 


.8124 


.9098 


20 


.9483 


.6254 


50 


.5854 


.7675 


.7221 


.8586 


1.3848 .1414 


.8107 


.9089 


10 


.9454 


.6283 


36° 00' 


.5878 


.7692 


.7265 


.8613 


1.-3764 .1387 


.8090 


.9080 


54° 00' 


.9425 






Value 


Logio 


Value 


Loffio 


Value Logio 


Value 


I^Ogio 


Degeees 


Radians 






Cosine 


Cotangent 


Tangent 


Sine 







542 Four Place Trigonometric functions 

[Characteristics of Logarithms omitted — determine by the usual rule from the value] 



Radians 


Deobeee 


8lNB 


Tangent 


Cotangent 


Cosine 










Value Logio 


Value Logio 


Value Logifl 


Value Logic 






.6283 


36° 00' 


.5878 .7692 


.7265 .8613 


1.3764 .1387 


.8090 .9080 


64° 00' 


.9425 


.6332 


10 


.5901 .7710 


.7310 .8639 


1.3680 .1361 


.8073 .9070 


60 


.9396 


.6341 


20 


.5925 .7727 


.7355 .86(J6 


1.3597 .1334 


.8056 .9061 


40 


.9367 


.6370 


30 


.5948 .7744 


.7400 .8692 


1.3514 .1308 


.8039 .9052 


30 


.9338 


.6400 


40 


.5972 .7761 


.7445 .8718 


1.3432 .1282 


.8021 .9042 


20 


.9308 


.6429 


50 


.5995 .7778 


.7490 .8745 


1.3351 .1255 


.8004 .9033 


10 


.9279 


.6458 


37° 00' 


.6018 .7795 


.7536 .8771 


1.3270 .1229 


.7986 .9023 


63° 00' 


.9250 


.6487 


10 


.6041 .7811 


.7581 .8797 


1.3190 .1203 


.79f39 .tX)14 


50 


.9221 


.6516 


20 


.()065 .7828 


.7627 .8824 


1.3111 .1176 


.7951 .9004 


40 


.9192 


.6545 


30 


.6088 .7844 


.7()73 .8850 


1.3032 .1150 


.7934 .8995 


30 


.9163 


.6574 


40 


.6111 .7861 


.7720 .8876 


1.2954 .1124 


.7916 .8985 


20 


.9134 


.6603 


50 


.6134 .7877 


.77(J6 .8902 


1.2876 .1098 


.7898 .8975 


10 


.9105 


.6632 


38° 00' 


.6157 .7893 


.7813 .8928 


1.2799 .1072 


.7880 .8965 


52° 00' 


.9076 


.6601 


10 


.6180 .7910 


.7860 .8954 


1.2723 .1046 


.7862 .8955 


50 


.9047 


.6u90 


20 


.6202 .7926 


.7907 .8980 


1.2647 .1020 


.7844 .8945 


40 


.9018 


.6720 


30 


.6225 .7941 


.79r>4 .9006 


1.2572 .0994 


.7826 .8935 


30 


.8988 


.6749 


40 


.6248 .7957 


.8002 .9032 


1.2497 .0968 


.7808 .8925 


20 


.8959 


.6778 


50 


.6271 .7973 


.8050 .9058 


1.2423 .0942 


.lim .8915 


10 


.8930 


.6807 


39° 00' 


.6293 .7989 


.8098 .9084 


1.2349 .0916 


.7771 .8905 


61° 00' 


.8901 


.6836 


10 


.6316 .8004 


.8146 .9110 


1.2276 .0890 


.7753 .8895 


50 


.8872 


.6865 


20 


.6338 .8020 


.8195 .9135 


1.2203 .0865 


.7735 .8884 


40 


.8843 


.6894 


30 


.6361 .8035 


.8243 .9161 


1.2131 .0839 


.7716 .8874 


30 


.8814 


.6923 


40 


.6383 .8050 


.8292 .9187 


1.2059 .0813 


.7698 .8864 


20 


.8785 


.6952 


60 


.6406 .8066 


.8342 .9212 


1.1988 .0788 


.7679 .8853 


10 


.8756 


.6981 


40° 00' 


.6428 .8081 


.8391 .9238 


1.1918 .0762 


.7660 .8843 


60° 00' 


.8727 


.7010 


. 10 


.6450 .8096 


.8441 .9264 


1.1847 .0736 


.7642 .8832 


50 


.8698 


.7039 


20 


.6472 .8111 


.8491 .9289 


1.1778 .0711 


.7623 .8821 


40 


.8668 


.7069 


30 


.6494 .8125 


.8541 .9315 


1.1708 .0685 


.7604 .8810 


30 


.8639 


.7098 


40 


.6517 .8140 


.8591 .9341 


1.1640 .0659 


.7585 .8800 


20 


.8610 


.7127 


50 


.6539 .8155 


.8642 .9366 


1.1571 .0634 


.7566 .8789 


10 


.8581 


.7156 


41° 00' 


.6561 .8169 


.8693 .9392 


1.1504 .0608 


.7547 .8778 


49° 00' 


.8552 


.7185 


10 


.6583 .8184 


.8744 .9417 


1.1436 .0583 


.7528 .8767 


50 


.8523 


.7214 


20 


.6604 .8198 


.87m .9443 


1.1369 .0557 


.7509 .8756 


40 


.8494 


.7243 


30 


.6626 .8213 


.8847 .94(38 


1.1303 .0532 


.7490 .8745 


30 


.8465 


.7272 


40 


.6648 .8227 


.8899 .9494 


1.1237 .050(3 


.7470 .8733 


20 


.8436 


.7301 


50 


.6670 .8241 


.8952 .9519 


1.1171 .0481 


.7451 .8722 


10 


.8407 


.7330 


42° 00' 


.6691 .8255 


.9004 .9544 


1.1106 .0456 


.7431 .8711 


48° 00' 


.8378 


.7359 


10 


.6713 .8269 


.9057 .9570 


1.1041 .0430 


.7412 .8699 


50 


.8348 


.7389 


20 


.6734 .8283 


.9110 .9595 


1.0977 .0405 


.7392 .8688 


40 


.8319 


.7418 


30 


.6756 .8297 


.91(33 .9()21 


1.0913 .0379 


.7373 .8676 


30 


.8290 


.7447 


40 


.6777 .8311 


.9217 .9646 


1.0850 .0354 


.7353 .8665 


20 


.8261 


.7476 


50 


.()799 .8324 


.9271 .9671 


1.0786 .0329 


.7333 .8(353 


10 


.8232 


.7505 


43° 00' 


.6820 .8338 


.9325 .9697 


1.0724 .0303 


.7314 .8641 


47°00' 


.8203 


.7534 


10 


.6841 .8351 


.9380 .9722 


1.0661 .0278 


.7294 .8629 


60 


.8174 


.7563 


20 


.6862 .8365 


.9435 .9747 


1.0599 .0253 


.7274 .8618 


40 


.8145 


.7592 


30 


.6884 .8378 


.9490 .9772 


1.0538 .0228 


.7254 .8606 


30 


.8116 


.7621 


40 


.6905 .8391 


.9545 .9798 


1.0477 .0202 


.7234 .8594 


20 


.8087 


.7650 


60 


.6926 .8405 


.9601 .9823 


1.0416 .0177 


.7214 .8582 


10 


.8058 


.7679 


44° 00' 


.6947 .8418 


.9657 .9848 


1.0355 .0152 


.7193 .8569 


46° 00' 


.8029 


.7709 


10 


.6967 .»i31 


.9713 .9874 


1.0295 .012(3 


.7173 .8567 


60 


.7999 


.7738 


20 


.6988 .8444 


.9770 .9899 


1.0235 .0101 


.7153 .8545 


40 


.7970 


.7767 


30 


.7009 .8457 


.9827 .9924 


1.0176 .0076 


.7133 .8532 


30 


.7941 


.7796 


40 


.7030 .84(59 


.9884 .9949 


1.0117 .0051 


.7112 .8520 


20 


.7912 


.7825 


50 


.7050 .8482 


.9942 .9975 


1.0058 .0025 


.7092 .8507 


10 


.7883 


.7864 


46° 00' 


.7071 .8495 


1.0000 .0000 


1.0000 .0000 


.7071 .849.-. 


45° 00' 


.7854 




' 


Value Logjo 


Value Lopio 


Value Logio 


Value Logio 


DSGSEEB 


Radians 






OOSINB 


Cotangent 


Tangent 


Sine 







INDEX 



Abscissa, 39. 

Absolute error, 236. 

Absolute value, of a directed quan- 
tity, 5 ; of a number, 36, 438. 

Addition, 41 ; graphic, 50 ; laws 
of, 52 ; of angles, 145 ; formulas, 
in trigonometry, 201 ; with 
rounded numbers, 238 ; of vectors, 
435. 

Algebra, fundamental theorem of, 
456. 

Algebraic functions, 143. 

Algebraic scale, 3, 5. 

Analytic geometry, 84. 

Analytic representation of a function, 
21. 

Angle, definitions of, 143 ; directed 
— s, 143, 306 ; measurement of, 
144, 188, 190; addition and 
subtraction of, 145 ; functions of, 
147, 168; between lines, 146 
306, 500; between planes, 506 
of elevation and depression, 150 
use of, in artillery service, 190 
vectorial, 163 ; — s of triangle, 
210; trisection of, 388; of a 
complex number, 438. 

Approximation, Newton's method of, 
471. 

Arbitrary functions, 18, 19. 

Arc of a circle, 189. 

Archimedes, 386. 

Area, of a triangle, 186, 298 fif. ; of 
any polygon, 301. 

Arithmetic mean, 214. 

Arithmetic progression, 216. 

Arithmetic scale, 3, 5. 

Artillery service, use of angles in, 
190. 

Asymptotes, 278, 280, 350. 

Axes, of reference, 38, 495 ; of ellipse, 
273, 340 ; of hyperbola, 280, 349. 



Axioms, 53. 

Axis, of parabola, 109, 354; polar, 
163 ; major and minor, 340 ; trans- 
verse, 280, 349 ; conjugate, 349. 

Binomial theorem, 428. 
Briggian logarithms, 227. 
Bundle of planes, 510. 

Cardioid, 383. 

Center, of pencil of lines, 91 ; of 
ellipse, 273, 282, 340; of circle, 
283, 320 ; of hyperbola, 280, 349 ; 
of projection, 370; of bundle of 
planes, 510. 

Change, rate of, 68. 

Change ratio, 69. 

Characteristic of a logarithm, 227 
ff. 

Circle, 320 ff . ; center and radius 
of, 320 ; cartesian equation of, 270, 
320 ff . ; parametric equations of, 
392 ; polar equation of, 381 ; as 
special case of an ellipse, 342 ; 
intersection of two — s, 330 ; or- 
thogonal — s, 331, 334; pencil of 
— s, 332 ; radical axis of, 332 ; radi- 
cal center of, 334 ; tangent to, point 
form, 324 ; slope form, 325 ; tan- 
gents from an external point to, 
327; polar of point with respect 
to, 328 ; inversion with respect 
to, 336. 

Cissoid, 384. 

Cof unction, 177. 

Colatitude, 530. 

CoUinearity, condition for, 301. 

Combinations, 420 ff . 

Commensurable segments, 33. 

Common logarithms, 227. 

Completing the square, 113, 283 ff. 



543 



544 



INDEX 



Complex numbers, 432 ff. ; defini- 
tion of, 432 ; geometric inter- 
pretation of, 433, 436 S. ; absolute 
value, angle, argument of, 438 ; 
polar form of, 438 ; multiplication 
and division of, 440; powers and 
roots of, 444. 

Complex roots of an equation, 462. 

Components of a vector, 436. 

Composite number, 414. 

Computation, numerical, 231, 236 ff., 
242 ff. 

Conchoid, 385. 

Cone, 528. 

Conic or Conic section, 337 ff. ; as 
sections of a cone, 370. (See 
circle, ellipse, parabola, hyperbola.) 

Conjugate axis, 349. 

Conjugate complex numbers, 432. 

Conjugate diameters, 376. 

Conjugate hyperbolas, 353. 

Consistent equations, 94, 491. 

Constant function, 18. 

Continuous functions, 18, 102, 449. 

Coordinates, on a line, 37 ; rectan- 
gular, in a plane, 38 ; rectangular, 
in space, 495 ; polar, in a plane, 
163, 377 ff. ; polar, in space, 530 ; 
spherical, 530; cylindrical, 531. 

Cosecant, 168; graph of, 173. 

Cosine, definition of, 147 ; variation 
of, 159 ; graph of, 159 ; graph 
of, in polar coordinates, 166 ; line 
representation of, 169 ; law of — s, 
180 ; direction — s of a line, 498. 

Cotangent, definition of, 168; graph 
of, 173; line representation of, 
169. 

Coversed sine, 168. 

Cube, duplication of, 388; table of 
— s, and — roots, 634. 

Cubic function, 129 ff. 

Cycloid, 399. 

Cylinders, 515. 

Cylindrical co5rdinates, 531. 

Decreasing function, 24. 
De Moivre's theorem, 443. 
Dependent variable, 12. 
Depressed equation, 469. 



Derivative, of a function, 451, 468 flF. ; 

successive — s, 458. 
Derived function, 451. 
Descartes's rule of signs, 466. 
Detached coefficients, 402 ff. 
Determinants, 475 ff. ; definition of, 

475, 478, 483 ; evaluation of, 488 ; 
properties of, 483 ff, ; minor of, 
485 ; Laplace's expansion of, 486 ; 
solution of equations by means of, 

476, 480, 490. 

Diameter, of a conic, 373 ; conjugate 
— s, 376. 

Diodes, 384. 

Directed angles, 143, 306. 

Directed lines, 30G, 500. 

Directed quantities, 4. 

Directed segments, 5, 48, 497. 

Direction angles and cosines, 498. 

Directrix, of conic, 337 ; of ellipse, 
340; of hyperbola, 350; of parab- 
ola, 355. 

Discontinuous functions, 18. 

Discriminant of a quadratic, 124. 

Distance, between two points in a 
plane, 294 ; in space, 499 ; of a 
point from a line, 311 ; of a point 
from a plane, 508. 

Division, by zero, 46 ; with rounded 
numbers, 241 ; of complex 
numbers, 440; point of, 295, 
501 ; — transformation, 404. 

Duplication of cube, 388. 

Eccentricity, 337. 

Element of a determinant, 475, 478. 

Ellipse, definition of, 272, 338 
axes and center of, 273, 340 
equations of, 272, 282, 338 ff. 
slope of, 273 ; construction of, 
346 ; focal radii of, 345 ; latus 
rectum of, 343 ; parametric equa- 
tions of, 347, 393 ; properties of, 
340, 342, 365, 373; vertices, 
343. 

Ellipsoid, 518 ff. 

Elliptic paraboloid, 526. 

Empirical function, 18. 

Equality, 51 ; conditional and un- 
conditional, 61. 



INDEX 



545 



Equation, definition of, 61 ; linear 
— s, 64, 83, 93 ff ., 509 ; quadratic, 
120 £f.; trigonometric — s, 174; 
exponential — s, 234 ; solution by 
determinants, 476, 480, 490; in 
p-form, 467 ; depressed, 469 ; 
parametric, 392 ff. 

Error, absolute and relative, 236. 

Explicit function, 81. 

Exponential equations, 234. 

Exponential function, 217 ff. 

Exponents, 53, 218. 

External secant, 168. 

Factor, 51 ; of a polynomial, 404, 
407; — theorem, 411. 

Factoring, solution of quadratic 
equation by, 121. 

Fire, indirect, in artillery service, 
190. 

Focal radii, of ellipse, 345 ; of hyper- 
bola, 353. 

Focus, of conic, 337 ; of ellipse, 340 ; 
of hyperbola, 350; of parabola, 
355. 

Forces, parallelogram of, 184, 435. 

Fractions, 33, 58 ; partial, 416 fif . 

Function, idea of, 1 ; definition of, 
28 ; arbitrary, constant, empirical, 
18; continuous, 18, 102, 449; 
representation of, 10, 13, 21, 22; 
increasing and decreasing, 24 ; 
linear, 64 ff., 494 ff. ; quadratic, 
98 ff., 265 ff., 514 ff. ; cubic, 129 ff. ; 
power, 140; trigonometric, 147 ff., 
168 ff.; logarithmic, 212 ff. ; ex- 
ponential, 217 ff. ; polynomial, 
449 ff. ; explicit and implicit, 81 ; 
inverse trigonometric, 192 ff. ; sum 
of linear — s, 79 ; tables of — s, 
534 ff. ; two- valued, 20, 265 ff. 

Functional notation, 409. 

Fundamental theorem of algebra, 
456. 

Geometric mean, 214. 
Geometric progression, 216. 
Geometric representation, see graphic 

representation. 
Graphic addition, 7, 50. 

2n 



Graphic interpolation, 13. 

Graphic representation, 3, 10, 37, 64, 
433, 436 ff. 

Graphic solution of problems, 78, 
123, 470. 

Graphs, statistical, 26 ; of linear 
functions, 72 ; of quadratic func- 
tions, 99 ff., 265 ff. ; of cubic 
functions, 129 ff. ; of polynomials, 
452 ; of trigonometric functions, 
158, 159, 161, 166, 173; of the 
exponential function, 221 ; of the 
logarithmic function, 224 ; in 
polar coordinates, 164, 378 ff. ; of 
parametric equations, 395. (.See 
entries under various classes of 
functions for further details.) 

Hesse's normal form of the equation 

of a straight line, 316. 
Highest common factor, 407. 
Hooke's law, 71. 
Hyperbola, definition of, 280, 338; 

axes and center, 280, 349 ; vertices 

of, 349; asymptotes of, 278, 

280, 355; construction of, 354; 

equations of, 279, 283, 348 ff.; 

parametric equations of, 393 ; 

latus rectum of, 350 ; geometric 

properties of, 349, 350, 368; 

tangents and normals to, 359; 

conjugate — s, 353. 
Hyperbolic curves, 140. 
Hyperbolic paraboloid, 527. 
Hyperbolic spiral, 386. 
Hyperboloid, of one sheet, 621 ; 

of two sheets, 524. 
Hypocycloid, 400. 

Identities, definition of, 61 ; trigono- 
metric, 171. 

Imaginary number, 432 ff. 

Implicit functions, 81 ; quadratic 
functions, 265 ff. 

Incommensurable quantities, 34. 

Increasing function, 24. 

Independent variable, 12. 

Infinity, 48. ■ 

Initial line, 163. 

Inscribed circle, 187. 



546 



INDEX 



Integer, 33 ; properties of, 414. 

Intercept, 74, 517 ; — form of equa- 
tion of straight line, 315. 

Interpolation, 13, 77, 230. 

Intersection, of two lines, 93 ; of 
two circles, 330; of a line and a 
conic, 357. 

Inverse trigonometric functions, 
192 ff. 

Inversion, with respect to a circle, 
336 ; of order, 482. 

Inversor of Peaucellier, 336. 

Irrational numbers, 34 ; as roots of 
an equation, 470. 

Laplace's expansion of a determi- 
nant, 486. 

Latus rectum, of an ellipse, 343 ; of a 
hyperbola, 350 ; of a parabola, 355. 

Law, of signs, 49, 466; — s of ex- 
ponents, 53, 220; of sines, 180/ 
of cosines, 180 ; of tangents, 209. 

Less than, 39. 

Limafon, 383. 

Line representation of trigonometric 
functions, 169. 

Linear equations, 64, 83, 93, 476, 
480, 490, 509. 

Linear functions, 64 fif., 494 ff. 

Linear interpolation, 77. 

Locus, of equation in rectangular 
coordinates, 67, 83, 509 ; in polar 
coordinates, 377. 

Logarithm, definition of, 223 ; in- 
vention of, 212; laws of, 225; 
systems of (natural and common), 
226, 227 ; characteristic and man- 
tissa of, 227 ff. ; use of tables of, 
229; — s in computation, 231, 
242 £F.; tables of, 536 ff. 

Logarithmic paper, 260 ff. 

Logarithmic scale, 252. 

Logarithmic spiral, 387. 

Longitude, 530. 

Magnitude, 4. 
Major axis of ellipse, 340. 
Mathematical analysis, 30. 
Maxima and Minima, 109, 137, 453, 
455. 



Measure, unit of, 2. 33, 34, 144, 188. 
Menelaus, theorem of, 319. 
Midpoint of a segment, 295, 502. 
MU, 190. 

Minor of a determinant, 485. 
Minor axis of ellipse, 340. 
Multiple roots, 460. 
Multiple-valued function, 20, 265 ff. 
Multiplication, 44, 50, 52, 239, 440. 

Napier, J., 212. 

Natural logarithms, 226. 

Newton's method of approximation, 
471 ff. 

Nicomedes, 385. 

Normal, to a curve, 359 ; — form of 
the equation of a straight line, 316 ; 
of a plane, 505. 

Number, 2, 33 ; real, 3, 36 ; rational 
and irrational, 34 ; positive and neg- 
ative, 3; complex (imaginary), 432 ff. 

Octant, 495. 

One-valued function, 20. 
Ordinate, 39. 
Origin, 38, 163, 495. 
Orthogonal circles, 331, 334. 
Orthogonal projection in space, 494. 

Parabola, 354 ff. ; definition of, 109, 
268, 338; equations of, 109, 282, 
354 ff.; properties of, 355, 362; 
slope of, 268 ; tangents and 
normals to, 359, 362 ; latus rectxim 
of, 365. 

Parabolic curves, 140. 

Parabolic reflector, 364. 

Parabolic spiral, 389. 

Paraboloid, of revolution, 364 ; 
elliptic, 526; hyperbolic, 527. 

Parallel lines, in plane, 85 ; in space, 
500. 

Parameter, definition of, 90, 392; 
of system of lines, 90. 

Parametric equations, 392 ff. 

Partial fractions, 416 ff. 

Pascal, B., 383,431 ; — 's triangle, 431. 

Peaucellier, inversor of, 336. 

Pencil, of lines, 91; of circles, 332; 
of planes, 510. 



INDEX 



547 



Period of trigonometric functions, 
157, 159, 162. 

Permutations, 420 ff. 

Perpendicular lines, in a plane, 86 ; 
in space, 500. 

Plane, 504 ff. ; vectors in a, 435. 

Point of division, in a plane, 295 ; 
in space, 501. 

Polar and Pole, with respect to a 
circle, 328 ; with respect to a 
conic, 364, 373. 

Polar axis, 163. 

Polar coordinates, in a plane, 163 ff., 
377 ff. ; in space, 530. 

Polar form of complex number, 
438. 

Polygon, area of, 301. 

Polynomials, 402, 449 ff. 

Power function, 140. 

Powers, 53, 218, 444 ; table of, 534. 

Prime number, 414. 

Principal diagonal of a determinant, 
475, 478. 

Principal value of inverse trigono- 
metric function, 193 ff. 

Probability, 426 ff. 

Product formulas in trigonometry, 
207. 

Progression, arithmetic and geomet- 
ric, 216. 

ProjectUe, 397 ff. 

Projecting cylinder of a curve, 516. 

Projection, 196 ff. ; central, 370; 
orthogonal in space, 494, 497. 

Pure imaginary number, 432. 

Quadrant, 39 ; angles in a, 145. 

Quadratic equation, 120 ff. 

Quadratic function, 98 ff. ; applica- 
tions of, 115 ff.; slope of, 274, 
287 ; implicit, 265 ff., 514 ff. 

Quantity, 2, 4 ; directed, 4. 

Radian, 188. 

Radical axis and center of two circles, 
332, 334. 

Radius vector, 163, 530. 

Range, of variable, 23; of a pro- 
jectile, 397. "■ 

Ratio, 33 ; simple, 294, 319. 



Rational numbers, 33 ; as roots of an 

equation, 467. 
Reciprocal spiral, 386. 
Rectangular coordinates, in a plane, 

38 ff.; in space, 495. 
Reference, axes of, 38. 
Reflector, parabolic, 364. 
Relative error, 236. 
Remainder theorem, 411. 
Revolution, surfaces of, 364, 519, 

520. 
Roots, of numbers, 444 ; table of, 

534 ; of an equation, 120, 455 ff. ; 

equal, of a quadratic, 124 ; rela- 
tion of, to coefficients, 125, 473 ; 

complex, 462 ; rational, 467 ; 

irrational, 470 ; multiple, 460 ; 

Newton's method of approximation 

to, 470 ff. 
Rotation in a plane, 200. 
Rounded numbers, 236 ff. 



Scale, arithmetic and algebraic, 
3, 5 ; rectilinear, uniform and non- 
uniform, 5 ; logarithmic, 252. 

Secant, definition of, 168 ; graph of, 
173 ; line representation of, 169. 

Section of a surface, 517. 

Segment, directed in a plane, 5, 48 ; 
in space, 497 ; — s to represent 
statistical data, 6. 

Shear, 132, 288. 

Significant figures, 236. 

Signs, law of, 49. 

Simple ratio, 294, 319. 

Simultaneous equations, 94 ff., 476, 
480, 490. 

Sine, definition of, 147 ; variation of, 
157 ; graph of, 158. 

Single-valued function, 20. 

Slide rule, 252 ff. 

Slope, 73, 135, 268, 273, 285, 287, 
290, 449. 

Solution, of quadratic equations, 
120 ff. ; of algebraic equations, 
468 ff. ; of trigonometric equations, 
174; of exponential equations, 

" 234 f of triangles, 181 ff., 244 ff. 

Sphere, 514. 



548 



INDEX 



Spherical co5rdinates, 530. 

Spirals, 386 ff. 

Statistical data and graphs, 6, 26. 

Straight line, 64 ff., 500 ff. ; slope 
of, 73; equations of, 83, 307 ff., 
511; intercept form of, 89, 315; 
normal form of, 315 ; parallel and 
perpendicular — s, 85, 500 ; system 
of — s, 90 ; intersection of — s, 93 ; 
polar equation of, 380 ; pencil of 
— s, 91 ; direction cosines and angles 
of, 498. 

Subnormal, 364. 

Successive derivatives, 458. 

Sum of linear functions, 79. 

Surd, 35. 

Surface, 504 ff., 514 ff. ; of revolu- 
tion, 364, 519 ff. 

Symmetric equations of straight 
line, 511. 

Synthetic division, 412. 

Table, of squares, etc., 118, 534 ; of 
logarithms, 536 ff. ; of trigono- 
metric functions, 538 ff. 

Tabular representation of a function, 
13. 

Tangent (to a curve), definition of, 
103 ; to a circle, 324 ff. ; to a 
conic, 360 ff.; slope of, 73, 107, 
135, 268, 273, 285, 287, 290, 449 ; 
slope forms of equation, 359 ; 
point forms of equation, 361. 

Tangent (trigonometry) , definition 
of, 147; variation of, 160; graph 



of, 161 ; line representation of, 
172 ; law of — s, 209. 

Taylor's theorem, 458. 

Term, 51. 

Trace of a surface, 517. 

Transverse axis of a hyperbola, 280, 
349. 

Triangle, area of, 186 ; angles of, 210 ; 
solution of, 181 ff., 244 ff. 

Trigonometric equations, 174. 

Trigonometric functions, definitions 
of, 147, 168 ; graphs of, 158, 159, 
161, 167, 173, 193, 194; variation 
of, 157 ff. ; computation of, 148 ff., 
152 ff. ; periods of, 157, 159, 162 ; 
inverse, 192 ff. ; formulas, 179, 180, 
181, 201, 202, 204, 205, 207, 446; 
application of De Moivre's 
theorem, to expansion of, 446 ; 
logarithms of, 242, 538 ff. ; tables 
of, 538 ff. 

Trisection of an angle, 388. 

Two-valued function, 20, 265 ff. 

Variable, definition of, 12 ; independ- 
ent and dependent, 12 ; range of, 
23. 

Vector, definition of, 5, 434 ; addition 
of — s, 435 ; components of, 
436. 

Vectorial angle, 163. 

Versed sine, 168. 

Vertex of a conic, 109, 343, 349. 

Zero of a function. 455. 



Printed in the United States of America. 



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TRIGONOMETRY 

BY 

ALFRED MONROE KENYON 

Professor of Mathematics in Purdue University 

And LOUIS INGOLD 

Assistant Professor of Mathematics in the University of 
Missouri 

Edited by Earle Raymond Hedrick 

Trigonometry, fie xible cloth, pocket size, long izmo {xi-\- ij2 pp.) with Complete 

Tables {xviii -\- 124 pp.) , $1.50 

Trigonometry {xi-{- ij2 pp.) with Brief Tables {xviii -\- 12 pp.), $1.20 

Macmillan Logarithmic and Trigonometric Tables, fiexlble cloth, pocket size, long 

i2mo (xviii + 124 pp.), $j6o 

FROM THE PREFACE 

The book contains a minimum of purely theoretical matter. Its entire 
organization is intended to give a clear view of the meaning and the imme- 
diate usefulness of Trigonometry. The proofs, however, are in a form that 
will not require essential revision in the courses that follow. . . . 

The number of exercises is very large, and the traditional monotony is" 
broken by illustrations from a variety of topics. Here, as well as in the text, 
the attempt is often made to lead the student to think for himself by giving 
suggestions rather than completed solutions or demonstrations. 

The text proper is short; what is there gained in space is used to make the 
tables very complete and usable. Attention is called particularly to the com- 
plete and handily arranged table of squares, square roots, cubes, etc.; by its 
use the Pythagorean theorem and the Cosine Law become practicable for 
actual computation. The use of the slide rule and of four-place tables is 
encouraged for problems that do not demand extreme accuracy. 

Only a few fundamental definitions and relations in Trigonometry need be 
memorized; these are here emphasized. The great body of principles and 
processes depends upon these fundamentals; these are presented in this book, 
as they should be retained, rather by emphasizing and dwelling upon that 
dependence. Otherwise, the subject can have no real educational value, nor 
indeed any permanent practical value. 



THE MACMILLAN COMPANY 

Fublishers 64-66 Fifth Avenue New York 



THE CALCULUS 

BY 

ELLERY WILLIAMS DAVIS 

Professor of Mathematics, the University of Nebraska 

Assisted by William Charles Brenke, Associate Professor ol 
Mathematics, the University of Nebraska 

Edited by Earle Raymond Hedrick 

Cloth, semi-Jlexible, xxi -\- 383 pp. + Tables {63), ismo, $2.10 
Edition De Luxe, flexible leather binding, India paper, $2. so 

This book presents as many and as varied applications of the Calculus 
as it is possible to do without venturing into technical fields whose subject 
matter is itself unknown and incomprehensible to the student, and without 
abandoning an orderly presentation i)f fundamental principles. 

The same general tendency has led to the treatment of topics with a view 
toward bringing out their essential usefulness. Rigorous forms of demonstra- 
tion are not insisted upon, especially where the precisely rigorous proofs 
. would be beyond the present grasp of the student. Rather the stress is laid 
upon the student's certain comprehension of that which is done, and his con- 
viction that the results obtained are both reasonable and useful. At the 
same time, an effort has been- made to avoid those grosser errors and actual 
misstatements of fact which have often offended the teacher in texts otherwise 
attractive and teachable. 

Purely destructive criticism and abandonment of coherent arrangement 
are just as dangerous as ultra-conservatism. This book attempts to preserve 
the essential features of the Calculus, to give the student a thcrough training 
in mathematical reasoning, to create in him a sure mathematical imagination, 
and to meet fairly the reasonable demand for enlivening and enriching the 
subject through applications at the expense of purely formal work that con* 
tains no essential principle. 



THE MACMILLAN COMPANY 

Publishers 64-66 Fifth Avenue New York 



GEOMETRY 

BY 
WALTER BURTON FORD 

Junior Professor of Mathematics in the University of 
Michigan 

And CHARLES AMMERMAN 

The William McKinley High School, St. Louis 

Edited by Earle Raymond Hedrick, Professor of Mathematics 

in the University of Missouri 

Plane and Solid Geometry, cloth, i2mo, 31 q pp., $1.23 
Plane Geometry, cloth, i2mo, 213 pp., $ .80 
Solid Geometry, cloth, i2mo, 106 pp., $ .80 

STRONG POINTS 

I. The authors and the editor are well qualified by training and experi- 
ence to prepare a textbook on Geometry. 

II. As treated in this book, geometry functions in the thought of the 
pupil. It means something because its practical applications are shown. 

III. The logical as well as the practical side of the subject is emphasized. 

IV. The arrangement of material is pedagogical. 

V. Basal theorems are printed in black-face type. 

VI. The book conforms to the recommendations of the National Com- 
mittee on the Teaching of Geometry. 

VII. Typography and binding are excellent. The latter is the reenforced 
tape binding that is characteristic of Macmillan textbooks. 

" Geometry is likely to remain primarily a cultural, rather than an informa- 
tion subject," say the authors in the preface. " But the intimate connection 
of geometry with human activities is evident upon every hand, and constitutes 
fully as much an integral part of the subject as does its older logical and 
scholastic aspect." This connection with human activities, this application 
of geometry to real human needs, is emphasized in a great variety of problems 
and constructions, so that theory and application are inseparably connected 
throughout the book. 

These illustrations and the many others contained in the book will be seen 
to cover a wider 7-ange than is usual, even in books that emphasize practical 
applications to a questionable extent. This results in a better appreciation 
of the significance of the subject on the part of the student, in that he gains a 
truer conception of the wide scope of its application. 

The logical as well as the practical side of the subject is emphasized. 
^ Definitions, arrangement, and method of treatment are logical. The defi- 
nitions are particularly simple, clear, and accurate. The traditional manner 
of presentation in a logical system is preserved, with due regard for practical 
applications. Proofs, both foimal and informal, are strictly logical. 



THE MACMILLAN COMPANY 

Publishers 64 G6 Fifth Avenue New York 



Analytic Geometry and Principles of Algebra 

BY 

ALEXANDER ZIWET 

Professor of Mathematics, the University of Michigan 

AND LOUIS ALLEN HOPKINS 

Instructor in Mathematics, the University of Michigan 

Edited by EARLE RAYMOND HEDRICK 

Cloth, via + j6g pp., appendix, answers, index, i2mo, $i.7S 

This work combines with analytic geometry a number of topics traditionally 
treated in college algebra that depend upon or are closely associated with 
geometric sensation. Through this combination it becomes possible to show 
the student more directly the meaning and the usefulness of these subjects. 

The idea of coordinates is so simple that it might (and perhaps should) be 
explained at the very beginning of the study of algebra and geometry. Real 
analytic geometry, however, begins only when the equation in two variables 
is interpreted as defining a locus. This idea must be introduced very gradu- 
ally, as it is difficult for the beginner to grasp. The familiar loci, straight 
line and circle, are therefore treated at great length. 

In the chapters on the conic sections only the most essential properties of 
these curves are given in the text ; thus, poles and polars are discussed only 
in connection with the circle. 

The treatment of solid analytic geometry follows the more usual lines. But, 
in view of the application to mechanics, the idea of the vector is given some 
prominence; and the representation of a function of two variables by contour 
lines as well as by a surface in space is explained and illustrated by practical 
examples. 

The exercises have been selected with great care in order not only to fur- 
nish sufficient material for practice in algebraic work but also to stimulate 
independent thinking and to point out the applications of the theory to con- 
crete problems. The number of exercises is sufficient to allow the instructor 
to make a choice. 

To reduce the course presented in this book to about half its extent, the 
parts of the text in small type, the chapters on soUd analytic geometry, and 
the more difficult problems throughout may be omitted. 



THE MACMILLAN COMPANY 

Publishers 64-66 Fifth Avenue New York 



Elements of Analytic Geometry 

BY 

ALEXANDER ZIWET 

Professor of Mathematics, the University of Michigan 

AND LOUIS ALLEN HOPKINS 

Instructor in Mathematics, the University of Michigan 

Edited by EARLE RAYMOND HEDRICK 

As in most colleges the course in analytic geometry is preceded by a course 
in advanced algebra, it appeared desirable to publish separately those parts 
of the authors' "Analytic Geometry and Principles of Algebra" which deal 
with analytic geometry, omitting the sections on algebra. This is done in the 
present work. 

In plane analytic geometry, the idea of function is introduced as early as 
possible; and curves of the form>' =/(x), where /(;c) is a simple polynomial, 
are discussed even before the conic sections are treated systematically. This 
makes it possible to introduce the idea of the derivative ; but the sections 
dealing with the derivative may be omitted. 

In the chapters on the conic sections only the most essential properties of 
these curves are given in the text ; thus, poles and polars are discussed only 
in connection with the circle. 

The treatment of solid analytic geometry follows more the usual lines. 
But, in view of the application to mechanics, the idea of the vector is given 
some prominence ; and the representation of a function of two variables by 
contour lines as well as by a surface in space is explained and illustrated by 
practical examples. 

The exercises have been selected with great care in order not only to fur- 
nish sufficient material for practice in algebraic work, but also to stimulate 
independent thinking and to point out the applications of the theory to con- 
crete problems. The number of exercises is sufficient to allow the instructor 
to make a choice. 



THE MACMILLAN COMPANY 

Publishers 64-66 Fifth Avenue New York 



SLIDE-RULE 



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A reasonably accurate slide-rule 
may be made by the student, for 
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Take three strips of heavy stiff 
cardboard 1".3 wide by &' long; 
these are shown in cross-section in 
(1), (2), (3) above. On (3) 
paste or glue the adjoining cut 
of the slide rule. Then cut strips 
(2) and (3) accurately along the 
lines marked. Paste or glue the 
pieces together as shown in (4) 
and (5). Then (5) forms the 
slide of the slide-rule, and it will 
fit in the groove in (4) if the work 
has been carefully done. Trim 
off the ends as shown in the large 
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14 DAY USE 

RETURN TO DESK FROM WHICH BORROWED 

LOAN DEPT. 

This book is due on the last date stamped below, or 

on the date to which renewed. 

Renewed books are subject to immediate recall. 



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